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(1,2 ss). The general equation of a line is y=mx+c, you can calculate the gradient (m), with the formula m=y2-y1/x2-x1. The coordinates (x1,y1) and (x2, y2) are specific point that are given. The equation of a line may be calculated with the formula: y-y1=m(x-x1). If y1=2, x1=3 and m=4: the formula to find the equation of a 2d line can be simplified first to y-2=4(x-3), then y=4x + 10.
The formula for the midpoint of a 2d line is (x1+x2/2, y1+y2/2) (1,2 ss).
(1,2 ss). You can find the length of a 2d line using Pythagoras' theorem. Where a=(x1, y1) and b=(x2, y2), the distance ab can be calculated with the formula: √((x2-x1)² + (y2-y1)²).
With simultaeneous equations, you can add and/or subtract [] and [] in the two seperate equations terms and answers--.
To find the point of intersection of straight lines, you can line up the equations and solve them like equations that are simultaneous. As both equations are equal to y, put them both as [] to one another equal. Then rearrange to find x [x is used in usual cases]. To find y, input the value(s) of x into any of the original equations.
When n appears more than once in a formula that seeks rearranging: seperate the terms depending on whether or not they include n. When you have the terms with n alone, make the side that n is on factorised with the term outside the brackets being n. To make n the subject of the formula, you may then divide by the terms that are multiplied by n.
y=3x+2, y=x²+3x-10, y=x^3 are all functions. In the three of these, the function... y is defined in terms of the variable x. If you change the value of x, then you change the value of y, thus y is 'dependent' on x. In all the functions, n(x) could replace y, but instead of n(x), often the notation that is used is f(x) (g(x) is also quite popular, but any letter can be used).
If f(x)=3x+3: f(2)= 9.
If y=f(x), then in the equation of a line, instead of y, it may be acceptable for one to use f(x).
If the equation of a graph is y= ax²+bx+c, where a and b and c are constants, then the graph is a parabola. H(x) is a quadratic, so the graph is a parabola, to find the roots solve: H(x)= 0. The roots are where the graph crosses the x-axis.
When the equation of a graph is y=1/ax+n, it is a transformation of the graph y= 1/x. To find the equation of the vertical asymptote, solve the equation: [] for [] ax+n=0 for x. If a=1: the equation of the horizontal asymptote is y=c. The asymptote is a line or curve, such that the distance between it and the graph approaches 0 as both tend towards infinity.
Let v=f(t), f(t)=2t 0=< t =<3, f(t)=6 3<t<9 and f(t)=24-2t 9 =< t =< 12. [#A29] From this: v=2t for values between and including 0 and 3(as it's a straight line: it's sufficient to just find f(t) for the line's limits). As f(t) = 6 where 3<t<9 - between t=3 to t=9: v= 6 and for values of t between 9 to 12 (inclusive): v= 24-2t. Again as it's a straight line it's sufficient to find f(t), just for the values of the line's limits.
For f(x)+g(x): add the two functions.
If a quadratic factorises to (x-p)(x+q), then the values of x that can be substituted into the quadratic for it to equal 0 are p or -q.
To display inequalities on a number line, first, you would mark off the edge(s) of the solution interval. This point might be marked with an unfilled circle around it or with an open parenthesis pointing in the direction of the rest of the solution interval. Then highlight the appropriate area. If instead of less than or more than, it's less than or equal to, or more than or equal to, then instead of a parenthesis, there is a square bracket and instead of an unfilled circle around a point, there's a filled (shaded) dot notation.
When solving inequalities, if you multiply or divide through by a negative, you then flip the inequality sign(s).
When trying to solve an inequality with 3 parts and 2 inequality signs (such as 2q<2r<2s): put the inequality into 2 parts, both sides containing the middle value. Solve them seperately, then put both parts together. (q<r, r<s: q<r<s.)
If you have the quadratic inequality: x²≤n, it can be simplified to -√n≤x≤√n. It has two limits as this often happens in a quadratic inequality.
The inequality: n²>9= n<3, n>3. 'This cannot be combined into one inequality!'
The places where the graph crosses the x-axis are called roots and the place where the graph crosses the y-axis is known as the y-intercept.
In a graph with the equation: y= ax²+bx+c, the y-intercept is c.
When trying to solve graphical inequalities graphically and there isn't (>, ≥, < or ≤)0, you can make it like that through rearrangement.
Different of two squares do not usually have x coefficients.
When factorising a quadratic that is the difference of two squares, in this case: n-o: the answer is (√n+√o)(√n-√o).
With the quadratic: (x-n)(x+o): the solutions for x are n and -o. It's like this as the quadratic equation equals 0, thus (x-n)(x+o) solves the equation if (x-n) and (x+o) equal 0.
{To complete the square for an quadratic in the form: x²+bx put the equation in the form (x±n)², where n is b * 0.5. This is known as making the square. Then expand the brackets and the difference between the factorisation and the original equation is , to the factorisation, either added or subtracted to make both equal. To do so for an equation in the form: x²+bx+c, do what was instructed for x²+bx, then add c.
{Once 'the square' has been completed, to solve the quadratic equation: rearrange for x, remembering that a number can have both positive and negative square roots, thus in (x+p)²-q, x can be -p±√q. When faced with a quadratic equation that does not include x², you may manipulate to get x².
{In an simultaneous equation(often quadratic) where n is on its own in one of the equations, substitute the value for n in one of the equation for n in the other equation(s).
When the equation of a circle is: (x-b)(y-c)=n, the radius of the circle is √n and the centre is (b,c).
When trying to find the coordinates of intersection with lines (and/or) curves and x or y = ±√n, the answer(s) do(es) not always use both, and moreover, somtimes neither values. Eg: when with the curves x^2 + y^2 = 16 and y = x^2 + 1; substitution and quadratic formula led to y giving 2 values, but all the points of intersection came from only one of the points (due to the root of negative numbers). It's good to [] values test-- and perhaps also check using a g[] graph. [I think that the case occurs where the product of the highest exponents of two functions is at least a quartic]
Polynomial terms have variables which are raised to positive integer exponents which are greater than 1. There are no square roots of variables, no fractional indices, and no variables should be used as a denominator. 4x² is a polynomial term. 6^4/2 is not a polynomial. √x is not a polynomial. 4/n² is not a polynomial. 3x²+3x-7 is
The coefficient on the leading term is known as the leading coefficient.
Any term that doesn't have a variable in it is called a constant term.
The degree of a term can be calculated from its exponent. A second-degree polynomial or a polynomial of degree two has an exponent of 2 and is sometimes called a quadratic. A fourth-degree polynomial has an exponent of 4 and is sometimes known as a quartic. A fifth-degree polynomial has an exponent of 5 and is sometimes known as a quintic.
The equation of a circle with centre (b,c) and radius r is: (x-b)²+(y-c)²=r².
To find the centre and the radius of a circle using its equation: put together the x and y terms. Change them seperately through factorisation. With factorisation: put the x-terms in the form: (x-b)²(+n if 'needed') and put the y-terms into the form: (y-c)²(+n if 'needed'). Then rearrange the equation into the form: (x-b)²+(y-c)²=r².
The equation of a circle is often in the form: (x-b)²+(y-c)²=r², where the centre of the circle is: b,c and the radius is r.
To find the place(s) where a line crosses the x-axis: input into the equation of the line: y=0, then find the root(s). To find the places where a line crosses the y-axis: input into the equation of the line: x=0, then find the y-intercept(s).
The tangent to a curve and the radius at that point are perpendicular.
To find the equation of a tangent to the circle with the equation (x-b)²+(y-c)²=r² at the point (n,o): using that the center of the circle is (b,c), find the gradient of the radius using points (b,c) and (n,o) with the formula m = y2-y1/x2-x1 (1,2 ss). The gradient of the radius is m, thus the gradient of the tangent is -1/m. As you know a point on the tangent and its gradient, calculate its equation with the formula (y-y1)=m(x-x1) (taking (n,o) to be (x1,y1) (1 ss).
A perpendicular bisector of a chord is the diameter, which goes through the centre of the circle.
To check if a coordinate is ON a line or curve: substitute its values for x and y into the equation of the graph. If the answer is the same as the answer in the original equation, then it's on the graph.
The perpendicular bisector of a chord should pass through the centre of both the circle and the chord.
Linear graphs are straight and have the equation: y=mx+c.
A graph with the equation: y=n/x is a reciprocal graph. The amount of sections it has is two. Often they are in quadrants that are diagonally opposite. If n is positive the lines of the graph take the quadrants which are the top-right and the buttom-left, but if n is negative, the graph is made []-[] upside down: the lines flip and cross the x-axis [#A30].
A graph with the equation: ax^3+bx²+cx+d is a cubic graph.
In a cubic graph: changing the value of the coefficient of y^3 ( in this case: a): by increasing the value of a: the graph is stretched in the []-axis y-axis by a scale factor of a. If a is negative: the graph becomes backwards. When a is equal to 0: the function becomes a quadratic.
By increasing the value of c(the y-intercept), the graph moves up the y-axis.
If the equation of a positive quadratic graph is changed into the form: (x+p)²+q, then the coordinates of the min point is (-p, q) and the y-intercept is p²+q.
If the translation of f(x)=x² is shown in the vector: (-p, q), then the equation of the line with the transformation, where y=f(x), is f(x)= (x+p)²+ q. Meaning to transform the graph f(x)=x², move it 'p' places to the left and q places up.
If a graph is in the form f(x)=(-)ax² + bx + c, more may be discovered about the graph by putting it in the form: f(x)= (x+[...] or [] (x+p)²+q or q-(x-p)².
With an equation in the form: f(x) = ax² + bx + c, to find the value of x that gives the lowest result: put the equation in the form: f(x)= (x+p)² + q, meaning that the answer is -p as the min point of the graph with equation f(x)=(x+p)² + q has coordinates (-p, q).
If the equation of a graph doesn't factorise into the form f(x)=(x+p)(x+q), it may be put into the form: f(x)=(x+p)²-q or q-(x-p)². To find the roots of the graph: change y or f(x) into 0 (as the roots of the graph are where the graph crosses the x-axis, which is where y= 0), after that rearrange for x, putting it in the form: -p±√q or p±√q (if the quadratic was negative). Some graphs although have no roots meaning that the equation has no solutions.
From the quadratic formula: if 'b²-4ac' gives a negative number, then the graph with that equation has no roots. That part of the equation is known as the discriminant. So to see if a graph has any roots input the appropriate values. The discriminant can be notated as Δ.
When a cubic equation factorises to (x-p)(x-q)(x-r): the roots are p, q and r. If a root appears more than once(such as: (x-p)(x-q)²) the graph touches q twice and goes through p. The point 'q' is an example of a repeated root. Quartic graphs can have 0 to 4 roots. The number that (x² + r)(x-p)² and (x-p)^4 have is 1. Graphs like (x-p)^4 may look like parabolas.
A quartic function often has a graph that looks like a 'w', but x^4 has a graph with the shape of a parabola.
In an equation where x is a variable, the other values in the brackets multiplied together equate to the constant. They are thus, of the constant, factors. So to find the roots of an equation: input into f(x), the values of the constant's factors and their n[] negatives. They are roots if the resultant is 0. If x=p is a root, then a factor of f(x)=0 is (x-p). As the numerical values in the brackets (not x) multiply to make the constant, if one or more constant is known (usually when only one factor is left), others may be deduced.
Cubic equation may be solved using the factor theorem. If all the potential integer roots have been checked and only one acceptable (x+p): then the other roots aren't integers. As the brackets multiply to create a cubic, the value that multiplies by the found factor to create the original equation is a quadratic. Put f(x) = (x+p)(x²+bx+c). Multiply the brackets, then to some parts factorise(x^3 + bx² + px²+ cx + pbx + c = x^3 +(b+p)x² + (c+pb)x + c). With the preceeding put equal to f(x), by inspection - find the value of both b and c. It is known that -p is a root, but so as to find the other roots: solve the quadratic.
When given a polynomial f(x), with an unknown value(a), and that divides by (x-p) to give n: f(p)= n. Put the equation that f(p) equates (which involves a) as equal to n, then find the value of a.
If f(p)=0, then a factor of the equation, with x as the variable is (x-p): this is called the factor theorem.
Where f(x) is an equation which can be factorised into (x+q)(x+r)(x+s): f(x)/(x+q)= (x+r)(x+s).
Dividing polynomial by linears. Division of x²-x-6 by x-3: x-3)¯x²-x-6. How many times does x-3 go into x²? x times, which equates to x²-3x. If this, from x²-x-6, is subtracted, the hence equates to 2x - 6. Divide the 2x-6 by x-3, to get 2. Using also that the answer to the first division was x: the answer to x-3)¯x²-x-6 is x+2. 2x²-x+5 divides by x-1 to 2x+1, remainder:6, thus 2x²-x+5=(x-1) (2x+1) + 6.
The remainder theorem states: any function f(x) can be expressed as the product of two other polynomials plus a remainder. Thus p(x) x q(x) + r= f(x). If r=0:, of f(x), p(x) and q(x) are factors. If a polynomial f(x) is divided by (ax-b), then the remainder is f(b/a), which is the answer. F(x)=x^3+nx²-3x-7 is divided by (x-2) with a remainder of 3. Find the value n. F(2)= 3. F(2)=2^3 + n(2²) + 3(-2) - 7 = 8 + 4n - 13 = 4n - 5 = 3: n= 2. (In some cases , esp. if there is more than one unknown and more than one value to be inputted into f(x), it may lead to a simultaenous equation).
An iterative or inductive sequence uses each term to find the next term. Usually what is given is both a formula and a starting number. Then the term(s) in the sequence are generated.
A sequence in which the difference of the successive values is a 'constant' is known as an arithmetic sequence. This 'constant'(d) is called the common difference.
When the terms in a sequence are added: the resulting sum is called a series.
The symbol used for sum of is Σ: the Greek letter sigma. On the symbol: it is often written, of the sum, the limits. At the bottom: there's k = n(not necessarily k), where n is the limit that's lowest, while at the top of the symbol: x, which is the limit given that's highest. If there's Σak(k sub-scripted), with x at the top and k=n at the bottom: it means add terms, where k has a value from n to x in ak(k sub-scripted). Without the ak(k sub-scripted) following the sigma: it would be the some of the numbers n to x.
Hypothetically: n is the amount of terms, a is the first term and d is the difference is. The first term in the sequence is a, the second: a+d, the third: a+2d and the last: a+(n-1)d. The sum of one pair is 2a+(n-1)d and as the amount of pairs is n/2: the formula for the sum to n terms in an arithmetic sequence is Sn (n ss) = n/2(2a+(n-1)d)
(x+1)^1=1x+1, (x+1)²=1x + 2x+1, (x+1)^3= 1x^3 + 3x² + 3x +1. The coefficients in the expressions can be placed into a triangle.(#L2#) In the 'triangle' each number (except for the first 1) is the sum of the two numbers above it. So from this the value of (x+1)^4 may easily be calculated. This 'triangle' of numbers is known as Pascal's triangle. 'These ideas can be applied to any expansion.' The values in the triangle remain. For (x+y)^n: use line [] of Pascal's triange n+1-- (e.g the 5th line would give 1, 4, 6, 4, 1). Multiply the numbers by the x-terms, with each x-term to the power of an indice, starting with n and for each next ones: the indice changing each time by -1. Multiply by the y-terms with the indice on the first one being 0 and the indice for the next ones changing each time by 1.
When (x+y)^n is expanded: the indices in each term add up to n.
The numbers in Pascals triangle are generated by the formula: ^nCr(r sub-scripted). 'n' represents the row and r, in that row, is the number's place. Also ^nCr(r sub-scripted) is the number of ways of choosing r objects from a group of n objects. The number of ways or the number in a position can be calculated with the formula: ^nCr(r sub-scripted)= n!/r!(n-r)!. The formula ^nCr(r subscripted) is sometimes written the form of the vector: (n r)(n above r in the vertical vector)or (n, r).
The binomial expansion of (x+y)^n is given by [characters immediately after 'C' subscripted]: ^nC0x^n + ^nC1x^(n-1)y + ^nC2x^(n-2)y²+...+ ^nCny^n. [#A31]. Also ^nCr(r ss) can be substituted for (n r) in a vertical vector. [#A32].
When ^nCr(r sub-scripted) is in referral to Pascal's triangle: n represents the row and r, in that row, is the number's place with the number in the first place having an 'r value' of 0 and the first row having an 'n' value of 0 - ^0Cr (r ss). And the first position on row n is given as ^nC0 (0 ss).
Calculating the gradient at any point on a curve is known as differentiation.
To find the gradient of a curve at (x, y): find the tangent, but as this can be difficult to do by hand, so an attempt at the estimation of the tangent can be made by instead looking at a nearby chord. This way, to get the line right, there can be a joining of 2 points. (#L2#.) In (-#L3#) the gradient can be found with the fraction: (3+h)²-9/h, which equals 6 + h, but when Q gets very close to p, h gets VERY close to 0. That technique can be applied to other values and polynomials. E.g if y = x^n and P=(x,y): Q would have coordinates x+h,(x+h)^n.
The general formula for the gradient of the tangent to a curve, where y=x²: as y=x²: the coordinate for the point on the curve is (x, x²) and as the x value of the other point on the tangent is bigger by the coordinate of the other crossed point on the (x+h, (x+h)²). The gradient can be put into the form of the fraction: (x+h)²-x²/h, which can be simplified into the form: 2x + h. As h gets very small: the gradient of the curve gets very close to 2x. With an x^3 graph it's different: the coordinates of the point is (x,x^3) and the coordinate of Q is (x+h,x+h^3), the gradient can thus be calculated from the fraction: (x+h)^3-x^3/h, which simplifies to 3x²+3xh+h². As h gets very, very small: the gradient of PQ gets very close to 3x².
The gradient of y=x^3 at general point: (x,y) is 3x², but if there's a coefficient(c) to x^3, then the gradient would be 3cx² . The gradient of y=x² at general point: (x,y) is 2x, but if there's a coefficient(c) to x², then the gradient would be 2cx. The coefficient - c, makes the line on the graph c times steeper. A y-intercept to the equation of the lines doesn't affect the gradient.
δn means an i... increment(i.e an infintesimal change) in n.
For differentiation: in general for y=x^n: the formula for the gradient is d[...] = [] dy/dx=nx^(n-1).
To differentiate y=c, where c is a numerical value: the answer is 0. To differentiate a polinomial with different indices included, differentiate each part seperately. For y=x^3-ox²+px-q, x^3 differentiated has the value 3x², -ox² has -2ox, px has p and -q has 0. If the different parts differentiated are 3x², -2ox, p and 0: y=x^3-ox²+px-q differentiated is 3x² - 2ox + p.
To find where y=f(x) differentiated equals n, find where dy/dx= n, differentiating f(x) gives an equation that's differential.
Another way of writing f(x) differentiated, using functional notation, is f'(x); other way are d y/dx (dy/dx) or y with above it: a dot. The notation dy/dx is the d[] derivative of y with respect to x.
When differentiating n/x, it may be changed to (with an indice) nx^-1.
The normal to a curve is a line that's s[] straight and to the tangent at that point is perpendicular, thus if the gradient of the tangent is g, the gradient of the normal is -1/g.
In the proof of the gradients of curves, using δ(to mean a small change in): P=(x,y), Q= (x+δx, y+δy). Thus dy/dx (the gradient formula) = lim δx→0 δy/δx.
If y=c(real number): dy/dx= 0. If y=f(x)+g(x): dy/dx= f'(x) +g'(x). If y=cf(x): dy/dx= cf'(x). To find the gradient at point 'a', substitute 'a' for 'x' in dy/dx.
To find the gradient of the tangent to the curve y=f(x) at the point where x=n: change f(x) by differentiation, and in f'(x) substitute x for n. The answer to f'(n) is (at that point) the tangent's gradient. Further: to find the equation of a tangent: using f(x) and x=n: find (at the point where x=n) the y-coordinate. As the gradient and a point on the line of the tangent are known, the equation of the line may be found using the formula: y-y1 = m(x-x1) (1 ss). Also, around the same may be done to the line that's perpendicular to the tangent (the normal).
The points on a graph where f'(x)=0 are called either stationary points or turning points. If f'(x) is negative before the point and positive after, the point is a minimum point. For a point to be a maximum point, f'(x) would have been positive then turned negative. If f'(x) remains as either negative or positive both before and after the point, then the stationary point is a point of inflexion.
The tangent at p on a curve goes in a positive direction if dy/dx> 0, in this part of the graph, as x increases: y increases. When dy/dx<0, the tangent goes in a direction that's negative, with respect to x: y is decreasing. Where the gradient(m)=0, the point has the name of (either) a stationary point or a turning point, the y-coordinate of the tangent has a stationary value and dy/dx = 0.
If the gradient of a tangent to a curve goes from positive to 0 to positive or negative to 0 to negative: where the gradient(m)=0 is the point of inflexion. To check if a point is a point of inflexion: substitute into dy/dx the values of x at, of the turning point, either side. If checking the values of x at either side of a point with a x value of one, instead of using 0 for checking, it may be better to use 0.5.
To find the value of x at turning points(where y=f(x)): find where dy/dx= 0, and then to find the y-coordinate, input the solutions of x into f(x).
The notation dy/dx is the rate that y changes as x changes. The differentiation of dy/dx is the second differential and may be used to find the rate that the gradient is changing. The second differential may be written as either d²y/dx² of f''(x) and it may be used to determine the nature of stationary points.
Where dy/dx<0: d²y/dx² is negative, where dy/dx>0: d²y/dx² is positive.
For stationary points dy/dx= 0.
One way to factorise cubics is to take out the common factor(s) and to the rest factorise.
To find the nature of a stationary point: to dy/dx differentiate, to get the second differential. Into the equation from the differentiation of dy/dx input, for each stationary point, the value of the x-coordinate. The second differential is less than 0 at a maximum point, the second differential is larger than 0 at a minimum point and 'if d²y/dx²=0, it could be anything.' To determine the nature in such a situation: on either side of the point find the gradient.
(A farmer needs a rectangular area of 36m² for his chickens. Find the lengths of the sides which use the smallest amount of fencing.) What should be small here is the perimeter. If it has width x(cm), then it has length 36/x(cm), thus the perimeter(L)=2*( 36/x + x)=72/x + 2x= (with an indice with the first fraction) 72x^-1 + 2x=L. For minimum perimeter (as it's a [] point): stationary point): 0= dL/dx= -72x^-2 + 2=0, thus x=6 (as the length isn't negative). One can check it's athe minimum using d²L/dx².
Where using maximum/minimum points or points of inflexion with applications, for example, if S=xy: rearrange so that there's only 1 variable.
Where dy/dx= p(a number)^1 +n, d²y/dx²= 0.
The formula for the gradient at any point is the gradient function (dy/dx).
The original function that when differentiated is equal to x^(a-1) + b is can be any of the equations in the form f(x)= x^a/a + bx + c, where c can be any constant, and is known as the [] constant arbitraty--. This is called a general solution and is the x^(a-1) + b after integration. The term that differentiates to give 4x is 2x². Integration involves using f'(x) to find f(x)(+c), which is known as the general solution.
If f'(x)=dy/dx=ax^n, y=f(x)= (ax^(n+1)/n+1) + c (when n is not equal to -1), which is equal to (with an integral symbol) ∫f'(x) dx. The value of c may be deduced for a specific graph, if a point on the specific graph is known (o,p) by putting o into f(x) and have it equating to p, thus p = f(o). E.g f'(x) = 2x, y = x²+c; if it passes through the point (1,9): 9=f(1)= 1+c, therefore c = 8.
Maybe we want to find the area under the curve f(x) between a and b. There is a small strip in the middle of the area, it is a narrow strip, so it's a small change in the area (δA). P is a point on the curve top-left corner, with coordinate (x,y) [#A33]: the strip is very narrow, of width δ x. Thus as the coordinate of the buttom-left corner is x, the x-coordinate of the buttom-right corner is x + δx. The top right corner has height y+δy and thus coordinates (x + δx, y + δy). [#A4#]. Two estimates from the areas come from the retangle with dimensions []*[] and the rectange with dimensions []*[] y*δx and the rectangle with dimensions δx*(y+δy), the area of the strip lies between the two areas [#A5]. The area of the stip can be expressed with inequalities as [], which simplifies to yδx<δA<δx(y+δy), which simplies to y<δA/δx<y+δy. 'We are making the strip very, very narrow, so δy and δx are getting very small.' So (lim δy→0 δx→0) δA/δx→ [] = [] dA/dx = y, therefore A= ∫y dx. Thus the area can be calculated by integrating the equation y=f(x).
Find the area under the curve y=3x² between x=0 and x=4, ∫y dx= ∫3x² dx= x^3 + c. 'To find c, use the fact that area = 0 when x = 0.' C is thus equal to 0 and S (area) = x^3. 'To find the area we need substitute x = 4'. S = 4^3 = 64... sq units.
If finding the area between two values of x, where none are 0: in this case c is 0 and we may find the area by finding the area under a curve for the values of x from 0 to b, and from that subtract the area under the curve for the values of x from 0 to a. When finding the area under a curve between a and b: it may be notated with an integral sign as ∫^ba f(x) dx (a ss) or (without an integral sign) [∫y]^ba (a ss). To solve: substitute both x= a and x= b and from x=b, subtract x=a.
To find the area between a line (g(x)) and curve (f(x): find the points of intersection (a and b, b>a). The sought area can expressed, depending on the situation, as either (∫b)a g(x) dx - (∫b)a f(x) dx (a ss) or (∫b)a f(x) dx - (∫b)a g(x) dx (a ss). Usually, the first is used with [] parabolas and second with [] parabolas positive--negative. Before integrating, it is possibly to subtract, giving either: (∫b)a f(x) dx - g(x) dx (a ss) or (∫b)a g(x) dx - f(x) dx (a ss). This may be solved through integration (though integration is not always needed to find the area between the line and the axis). This can also work when there's an enclosed area under the x-axis. The technique is similar for two curves: the enclosed area between f(x) and g(x), where f(x) overarchs the curve g(x) can be expressed as (∫b)a f(x) dx - g(x) dx (a ss).
To find the vector to move a basic x² graph onto an appropriate quadratic graph: put the equation of the quadratic graph in the form of either (x + p)² + q or q - (x - p)².
A geometric sequence has its first term notated as 'a', the ratio of the terms is written as 'r'. The nth term can be found with the expession a x r^n-1.
A sequence in which the terms increment in an unchanging ratio is a geometric sequence.
When displaying inequalities on numberlines: if incorporating >/< x - above x on the numberline - place an [] [] unfilled circle and draw a line in the appropriate direction, stopping the line at an appropriate time. If => or =< is used, the circle should be filled.
Expressions with inequalities are 'always' written from [] number smallest number to biggest.
Common factors don't have to appear in the common denominator more than once. Eg: to simplify a/bn + x/yn: the common denominator could be written as bny. The numerator a in the first fraction would be multiplied by y and the numerator x in the second fraction would be multiplied by b.
Before we multiply or divide fractions, diagonally as well as vertically, common factors can be cancelled. Eg: 2/15 * 3/8 simplifies to 1/15 * 3/4, which simplifies to 1/5 * 1/4.
To find the length of a tangent from the point (x,y) to the circle f(x) (where the equation is known): as the tangents meet radii at 90 degrees, the length can be found through the use of [] theorem Pythagoras' theorem. To do so, you may find the circle's center.
In graphs if y=√x, the graph is as the line y=x^2, but rests on the line x=0 and y only takes the square root values that are positive.
The graph of y = 1/x² has two lines, both in different quadrants; it is alike the graph of y=1/x except both lines are in quadrants above the x-axis and the lines are to the axes even closer.
If f(x) = 1/x: 1/(x-2) = f( x-2), which would move the graph y=1/x two places to the right on the x-axis.
If the equation of a negative quadratic graph is changed into the form: q-(x-p)² , then the coordinates of the max point is (p, q) and the y-intercept is q-p². E.g y=4-(x-2)², with max point (2,4) and y-intercept 0.
In the quadratic formula 'b²-4ac' (Δ) is called the discriminant. If b²-4ac>0, the amount of roots is two--, if b²-4ac=0, the amount of roots is one and a quadratic equation has no roots if the discriminant is less than 0 (b²-4ac<0).
The amount of turning points that quartic functions can have is either 1 or 3.
If a polynomial has degree n, then the maximum number of stationary points it has is n-1.
To solve an equation with an unknown power, rewrite it so that both sides have the same base.
With complex equations such as 3^2x - 10(3^x) + 9 = 0: y can be used as a substitute for 3^x. This creates and equation that's a quadratic: y^2-10y+9 = 0, which has solutions y = 1,9, thus 3^x = 1,9, therefore x = 0,2.
With 'r' as the ratio between numbers, the general formula for geometric sequences is ak+1 (k+1 ss) = ak*r (k ss)
A formula that can be used to generate sequences directly (e.g ak [k ss] = 2k + 1) is a deductive formula (also known as a direct formula).
For an arithmetic sequence with first term 'a' and common difference 'd': the nth term - un(nss) = a + (n-1)d.
If an arithmetic sequnce has first term 'a', common difference 'x' (unknown) and nth term 7: put a + (n-1)x = 7 and solve for x. (This technique can be applied to other situations, including those where other components [e.g 'a'] are missing).
The nth term for triangle numbers has the same formula as the formula for the sum of n n[] numbers natural--.
With (a+b)^n, the indices attached to each cooeficient add to n.
If y=f(x) the derivative of y w.r.t x is given by: dy/dx= (lim δx→0) f(x+ δx)-f(x) / δx
To find the coordinates of the points on the graph where the graph has gradient m, put f'(x) = m, then solve for x, before inputting the solution value(s) of x into f(x) to find the values of y. E.g to find where f(x) is parallel to y=4-x: put f'(x)=-1.
With application of minimum or maximum (and maybe even inflexion points), make sure that each side of the equation contains only one v[] variable. If there are more than one variables on one side (e.g A=2xy), (a) variable(s) can be replaced through substitution.
The area under a speed/time graph represents the distance travelled.
{x | x is a real number, x < 3} and ((#a1)). The pronunciation (what it means) is that: the set of all x, such that x is (both) a real number and x is less than minus three.{x | x < -3}. The pronunciation of this is: all x such that x is less than minus three. Such notation is set notation. The pronunciation for(-∞, -3) is the i[] from ... interval from -∞ to -3. Those were different notations for expressing the inequality: x < -3.
Created by: Toluo