gradient and graphs

how to find the mid point between 2 sets of coordinates along a straight line?

((x1 + x2)/2 , (y1 + y2)/2)

in a graph, the turning point is (5, 2). what are the coordinates of the new position of the turning point when the graph y = f(x) is transformed to the graph of: y = - f(x) y = f(-x) y = f(x) + 2 y = f(x+3)

(5, -2) (-5, 2) (5, 4) (2, 2)

how to find the mid point?

(a + b)/2

solve this simultaneous equation algebraically 2x + 5y = 18 4x - 3y = 10

(multiply the first equation by 2) y = 2 x = 8

write x^2 + 6x -5 in completed square form

(x + 3)^2 - 14

write x^2 + 7x + 14 in completed square form. leave decimals as improper fractions.

(x + 7/2)^2 + 7/4

to write the quadratic x^2 + bx + c in completed square form, you need to make it loo like:

(x + p)^2 + q

quadratic equation: 0 = x^2 - 3 what is x when y = 0?

+/- root 3

how do you solve: x^2 - 5x - 6 > 0

- 1 > x > 6 - factorise the quadratic equation and plot onto a graph. shade the areas where y is greater than zero on the graph. greater than sign etc can act like an equals sign, so values on the right hand side can be moved across and change positive/negative

Line A passes through the points (-3,-1) and (-1,9) Line B passes through points (-2,1) and (k,4) Line A and B are perpendicular. Find the value of K

- 17

Line A passes through the points (3,6) and (5,-2) Line B passes through points (2,5) and (8,k) Line A and B are parallel Find the value of K

- 19

how to find the equation of a straight line from 2 points

- find the gradient - find the point slope formula y - y1 = m(x - x1) - then substitute in values so that y is the subject of the formula.

how to solve a missing value of a coordinate when given 2 whole coordinates that are all on a straight line e.g. (1, 6) (5, -2) and (a, -12) are all on a straight line. Find A

- find the gradient - then substitute x and y vales with one coordinate that is already known into equation with gradient filled in too - solve y intercept - then rewrite the equation where x and y are unknown and fill in the intercept and gradient - then substitute only known vaue of mussing coordinate into equation and solve so that x or y is the subject of the formula.

method to write equation in square form:

- first writ the sqaure bracket. the value of p is always b/2. so the square bracket would be (x + b/2)^2 - expand bracket to get x^2 + bx + (b/2)^2 - finally find the amount you need to get from (b/2)^2 to c in the original equation. the difference needed is q.

solve each of the following equations graphically: x^2 - 3x -3 = 0 x^2 + 8x + 5 = 0 2x^2 + 3x - 6 = 0

-0.8, 3.8, -7.3, -0.7 -2.6, -1.1??????????????

the product of the gradients of 2 perpendicular lines is...

-1 (use this fact to compare 2 lines when they are each rearranged in the equation to see if they are perpendicular with each other)

how many possible values are there to find of x on a cubic graph?

1 or 2 or 3

solve: x^2 + 3x - 4 ≤ 0 x^2 - 9 ≥ 0 x^2 + 8x + 12 < 0 x^2 - 3x - 10 ≥ 0 (x+3)^2 ≤ 16 4x^2 ≥ 9 y^2 + y > 12 16 - x^2 > 0 (x + 1)^2 + (x+2) < 25 10 - (2x+1)^2 ≥ x

1 ≥ x ≥ - 4 - 3 ≥ x ≥ 3 - 2 > x > - 6 - 2 ≥ x ≥ 5 1 ≥ x ≥ - 7 - 1.5 ≥ x ≥ 1.5 - 4 > y > 3 - 4 > x > 4 -5 < x < 2 -9/4 ≤ x ≤ 1

how do you calculate the area of a trapezium?

1/2 (a + b) x height

factorise and solve x in the equation: 10x^2 + 9 = 21x

10x^2 -21x + 9 = 0 !0 x 9 = 90 10x^2 - 6x - 15x + 9 = 0 2x (5x - 3) -3(5x - 3) = 0 (2x - 3)(5x - 3) = 0 x = 3/2 x = 3/5

you can use this formula to change a temperature C in degrees c into a temperature F in degrees f. F = 1.8C + 32 use this to change 20 degrees c into degrees f

60 degrees f

The height of a ball above the ground in metres can be calculated from the formula h = 30t - 5t^2 on a graph where height is along the y axis and time is along the x axis. find he total time? The maximum height? Time when height is 25cm?

6s 45cm 5s

find the equation of the line with a gradient of 3 which passes through (2,-1)

7 = 3x - 7

expand and simplify (x+2)(2x-3)(4x-1) (5x+2)(3x-3)(-5x-2) (y+1)^2 (y+3)

8x^3 + 6x^2 - 23x - 6 -75x^3 + 15x^2 + 18x + 12 y^3 + 5y^2 + 7y + 3

find the length of the line with end points (-3,2) and (0,-2)

= 4 (use Pythagoras' theorem)

how can the discriminant of quadratics vary depending on positive or negative values.

A positive discriminant indicates that the quadratic has two distinct real number solutions. A discriminant of zero indicates that the quadratic has a repeated real number solution. A negative discriminant indicates that neither of the solutions are real numbers.

does the point (-3,5) lie on the circle x^2 + y^2 = 36

No -3^2 + 5^2 = 34

A is the point with coordinates (1, 3) B is the point with coordinates (4, -1) The straight line L goes through both A and B. Is the line with equation 2y = 3x - 4 perpendicular to line L?

No because the gradient is not the negative reciprocal of the straight line equation.

dynamic equilibrium on a graph

When the line on the graph always eventually comes back to the same place

(1, 6) (5, -2) and (a, -12) are all on a straight line. Find A

a = 10

given that: (ax+1)(x-3)(x+b) = 2x^3 - 3x^2 - 8x - 3 find a and b

a = 2 b = 1

if (ax + b)^2 (x+c) = 4x^3 + dx^2 - 55x - 100 find integers of a, b , c and d

a = 2 b = 5 c = -11 d = 4 very very hard. try working out in alphabetical order. probs can't work out.

given that: (x+3)(x+a)(x+7) find the value of a

a = 5

what is the curved, symmetrical shape formed from a quadratic sequence called?

a parabola

indirect/inverse relationship on a graph

as one variable increases, the other decreases

what is the discriminant in the quadratic equation

b^2 - 4ac

quick tip for describing the progression of a graph - e.g. velocity time graphs

be systematic and go step by step from beginning to end in detail.

velocity time graphs: diagonally up = horizontal = diagonally down = gradient = area under the graph =

constant acceleration constant velocity constant deceleration the acceleration of the object the distance travelled by the object

y = -x on a graph

diagonal line from top left to bottom right

how to represent bounds on a graph

dotted line if the inequality is > or < and use a solid line if the inequality is ≤ or ≥ plot the straight line. e.g. if the inequality is x>2 then draw on the line x=2, or if it is y<x+1 then draw line x+1. shade the area the you don't want. to check this - plug in certain values into the inequality by finding a random coordinate within the range and seeing if it satisfies the inequality (not this does not include marks on the line you drew originally) tip - if the x value is positive, the inequality symbol of y is generally pointing in the direction you should shade and don't need, however this doesn't always work (e.g. negatives)

in a cubic graph, if the coefficient of x is negative the curve goes...

down from the top left

solving quadratics graphically

draw a table, of 2 rows. x goes on the top (e.g. values from -5 to 5) then substitute the line equation with the different x values to solve the y row. you can find the estimates of the roots of f(x) by plotting the graph and finding the x values when y is zero, along the x axis. (this only applies if the line equation is equal to zero) if the line equation is not equal to zero: e.g. if line equation is equal to 4 then draw line y = 4 on the graph. find the roots on this new line.

shading inequality regions on graphs

draw the line represented by making the inequality equal to x or y. if it is equal to x, the line will be vertical. if it is equal to y, the line will be horizontal. if it is equal to a linear equation, draw the line onto the graph. use dotted line if it is < or > use a solid line if it is ≤ or ≥ shade/label the region where the points satisfy the inequality symbol. see maths genie for questions

how to solve nth term, quadratic sequence

find difference between terms. find second difference between terms. divide the second difference by 2. this is the coefficient of n^2. then write out equation an^2. take this away from the original sequence given. then solve this new sequence as a normal linear nth term sequence and add the to an^2 value.

how to find the turning point of a parabola without using a graph

find the midpoint between the 2 x axis coordinates. midpoint = a + b / 2 put this x coordinate into the line equation and solve y. this gives the full coordinate for the turning point.

how o find the y intercept

find the y intercept by putting x=0 into the line equation.

exponential graphs

have the form y = a^x if a is greater than 0 they have a horizontal asymptote ay y = 0 and a y intercept at (0,1) as y = a^0 = 1 the bigger the value of A the steeper the graph

parallel lines

have the same gradient but their y axis varies

the graph of y = f(ax) is a....

horizontal stretch of the graph y = f(x) with scale factor 1/a, parallel to the axis.

how do you tell if a group of points sit on the same straight line on a graph?

if they have the same gradient and y intercept

the graphs of quadratic functions are always...

in the shape called a parabola and are symmetrical

perpendicular lines

intersect to form a right angle if the gradient of a line is m, then the gradient of the perpendicular line is - 1/m

what is different about the line equation for a quadratic graph than a linear graph?

it contains an x squared. however the power of x is never greater than 2.

if (x+k)^2 (x+2) = x^3 + 14x^2 + 60x + 72 find the value of k

k = 6

Asymptotes

lines which the graph approaches but never reaches - e.g. the axis in a reciprocal graph

what is velocity measured in?

m/s

What is acceleration measured in?

m/s^2

how to solve x in a quadratic equation?

make sure it is in the form so that the equation = 0 = y then factorise the equation make x on its own by passing the numbers across the equals sign. you will always get possible answers.

how to find the x intercept from a quadratic equation

make the equation equal to y as 0. then factorize it. this gives you (x + a)(x + b) = 0. therefore when you solve this, the values of x will be -a and -b. so the line will cross the axis at (-a, 0) and (-b, 0).

what are the 2 types of parabola?

minimum and maximum

a line sloping downwards from left to right has a...

negative gradient

perpendicular lines definition

non-vertical perpendicular lines have negative reciprocal gradients. The product of the gradients will be -1

if: (x+p)(x+q)(x+5) = x^3 + 8x^2 - 3x - 90 find values of p and q

p = -3 q = 6

how do you solve simultaneous equations graphically

plot both lines on the graph. find where the lines meet at x and y values. if there are 2 straight line graphs there will be one solution. if one is a quadratic equation, there will be four solutions.

a line sloping upwards from left to right has a...

positive gradient

the graph of y = - f(x) is the...

reflection of the graph of y = f(x) in the x axis.

the graph of y = f(-x) is the

reflection of the graph y = f(x) in the y axis

Describe the transformation that maps the curve with equation y = sin(x) onto the curve with equation: y = - sin (x) y = 1 + sin(x) y = sin(x - 30degrees)

reflection with mirror line of the x axis translation by vector (0 over 1) translation by vector (30 over 0)

important notes when solving gradient or straight lines

remember m and c in the equation can have a negative symbol beforehand make sure you rearrange the line equation

how do you calculate the acceleration of an object?

rise/run (calculate the gradient)

how do you layout your workings out for simultaneous equations

set it out like a normal addition or subtraction working out. with one equation on top of the other and two parallel, horizontal lines underneath.

the bigger the value of A in an exponential graph, the

steeper the graph

y = x on a graph

straight diagonal line from bottom left to top right

how to sketch a straight line equation on a graph:

substitute x for random values within the scale and plot. join all of them up making sure the line crosses through the y axis at the correct intercept.

how to find the y intercept from a quadratic equation

substitute x=0 into the equation and solve it.

the parabola is n shaped if...

the coefficient of x^2 is negative

the Parabola is u shaped if...

the coefficient of x^2 is positive

circle graphs

the equation of a circle with radius r ad centre (0,0) is x^2 + y^2 = r^2 to draw a circle from this equation, take the square root to find the value of r. then draw the circle through r on both sides using a pair of compasses with the centre being the origin.

how can you tell if a line equation is cubic?

the highest power of x is 3

The turning point

the highest/lowest part of the curve depending on whether or not it makes an n or a u shape. As quadratic graphs are symmetrical, the x coordinate of this point is always the x midpoint.

a line passes through the origin if it...

the line equation only contains x and y terms (e.g. y=5x NOT y=5x+4)

what features should you include when describing a parabola?

the line of symmetry the y intercept turning point quadratic line equation

how to find the turning point of a quadratic graph

the parabola is always symmetrical so the you need to find the midpoint of x roots to find the x coordinate. then substitute this into the line equation to find the y coordinate.

if the coefficient of x squared is negative then:

the parabola is n-shaped

if the coefficient of the x squared is positive then:

the parabola is u-shaped

cyclic relationship on a graph

the pattern on the graph repeats itself over and over again

what points are necessary to find in order to draw a quadratic graph?

the y intercept the x intercepts (remember there is 2) the turning point

what is important to remember when finding the value of x at a certain y value on a quadratic graph?

there must be 2 values as a parabola is symmetric. look at the graph to see if the value is actually a possible value and the coordinates are included in the curve.

how to sketch cubic graphs

they are normally three marks -shape -roots -y intercept factorise the equation if necessary to find the possible roots. remember the plus or minus may change. to find shape, work out if coefficient for x^3 value is positive or negative. find the y intercept by multiplying all of the coefficients (the possible x values). e.g. if factorised line equation was (x + 2) (x + 3) (x + 4) the y intercept would be 2 x 3 x 4 = 24 or you can expand the factorised equation and find the d value.

reciprocal graphs

they are undefined when x=0 and y=0. Therefore they have asymptotes. e.g. y = 1/x or y = -1/x

how many times do cubic graphs cross the y axis

they only ever cross the y axis once

the graph of y = f(x) + a is the......

translation of the graph of y = f(x) by the vector (0 over a)

the graph of y = f(x+a) is the........

translation of the graph of y = f(x) by the vector (-a over 0)

in a cubic graph, if the coefficient of x is positive the curve goes...

up from the bottom left

the graph of 2x^2 - 4x + 1 = k has exactly one solution, use the graph to find the value of k.

use a table to substitute x values for y until you find the turning point (because there is only one solution) the y value of the turning point is equal to k. k = -1

how to find the length of a line segment

use line as the hypotenuse and join the rest to make a right angled triangle. Use Pythagoras' theorem A^2 = B^2 + C^2

equation for a circle on a graph

using Pythagoras' theorem, the equation of a circle with radius r and centre (0,0) is given by the formula: x^2 + y^2 = r^2

equation for gradient

vertical distance / horizontal distance = change in x / change in y (left to right and low to high)

the graph of y = af (x) is a......

vertical stretch of the graph of y = f(x), with the scale factor a, parallel to the y axis.

the ac method

when factorising, if the x^2 has a number before it: if every quadratic is in the form ax^2 + bx +c then b has to add up to a multiplied by c then use the split method to create two brackets.

direct relationship on a graph

when one variable increases, so does the other

cubic graph - touches x axis once, crosses x axis once

when the graph of a cubic function y crosses the x axis once and touches the x axis once, the equation y = 0 has three solutions but one is repeated

cubic graph - crosses x axis once

when the graph of a cubic function y crosses the x axis one, the equation y=0 have one distinct, repeated solution. e.g. y = (x-1)^3 OR only one real solution (e.g. if it is above the x axis. this happens if the discriminant is negative. e.g. in the cubic equation (x+2)(x^2 + x + 1) the quadratic (x+2)(x^2 + x + 1) has no real solutions as the discriminant is negative.

cubic graph with 3 roots

when the graph of a cubic function y, crosses the x axis three times, the equation y = 0 has three solutions.

what is the turning point of a parabola?

where the curve dips or rises completely before mirroring the same shape on the other side. it is the lowest/highest point on the graph.

represent the following inequality on a graph x + y < 4

x + y < 4 can be rearranged to y < 4 - x so plot the straight line - x + 4 with a dotted line. and shade the top right diagonal corner of graph as this is not needed.

general equation for vertical line

x = b

midpoint =

x1+ x2 / 2

how to find the midpoint of a line segment

x2 - x1 / y2 - y1

draw the line of y = x^2 - 4 then use this graph to find the solutions to the equation x^2 - 2 = 0

x^2 - 2 becomes x^2 - 4 = - 2 x = -1 x = 2 x = -3 y = 0

draw the graph of y = x^2 - 2x - 2 use the graph to help you solve the following x^2 - 2x = 2 x^2 - 3x - 1 = 0 x^2 - 3 = 0

x^2 - 2x - 2 = 0 x = - 0.8 x = 2.8 x^2 - 2x - 2 = x - 1 x = 3.2 x = -0.4 x^2 - 2x - 2 = -2x + 1 x = 1.8 x = -0.8 (this can be confusing - e.g. negatives are positives etc>) make the original parabola equation equal to something that can then be rearranged to the new straight line graph.

use x^2 - 3x -2 = y to solve x^2 - 3x -8 = 0 to 1 d.p.

x^2 - 3x - 2 = 6 then draw along line y=6 then bring down the points where y=6 crosses the parabola to find both of the x values.

draw graph of y=x^2 - 4 use the graph to help you solve the following x^2 - 2 = 0 x^2 - x - 1 = 0 x^2 + 2x - 2 = 0

x^2 - 4 = - 2 x = -1.5 x = 1.5 x^2 - 4 = x - 3 x = - 0.5 x = 1.5 x^2 - 4 = - 2x - 2 x = 0.8 x = 2.8 (this can be confusing - e.g. negatives are positives etc>) make the original parabola equation equal to something that can then be rearranged to the new straight line graph.

use x^2 -4x -2 = y to solve x^2 -3x -8 = 0

x^2 - 4x - 2 = 6 -x draw this line on the graph and see what x values cross over the parabola

use y = x^2 - 4x -2 to solve x^2 -5x -3 = 0

x^2 - 4x -2 = x + 1 draw the line x+1 on the graph find the x values it crosses the parabola. when moving values across the equals symbol, work out how you get FROM the ORIGINAL equation TO the NEW one.

draw the graph of y = x^2 - x + 2 use the graph to help you solve the following x^2 - x = 0 x^2 - x - 4 = 0

x^2 - x + 2 = 2 x = 0 x = 1 x^2 - x + 2 = 6 x = -1.5 x = 2.5 Always write the line equation that is draw in front of you already first. then add to the other side of equal sign whatever is necessary to reach the new equation. (this can be confusing - e.g. negatives are positives etc>) make the original parabola equation equal to something that can then be rearranged to the new straight line graph.

quadratic functions always have...

x^2 as the highest power of x

find the equation of the straight line perpendicular to y = 5x - 1 through point (10, 3)

y = - 1/5 x + 5

find the equation of the line parallel to 2x + 5y = 10 which passes through (0, -3)

y = - 2/5 x - 3

Find the equation of the line that passes through the point (-1, 5) and is perpendicular to y = 1/3 x - 2

y = - 3x + 2

find the equation of the line perpendicular to 5y = 2x - 4 which passes through (0,7)

y = - 5/2 x + 7

find the equation of the line perpendicular to 5y = 2x - 4 which passes through (0, 7)

y = - 5/2 x + 7

what are 2 examples of reflections of y = f(x) in trig graphs

y = - f(x) (mirror line is the x axis) y = f(-x) (mirror line is the y axis)

Find the equation of the line with gradient -2 which passes through (-2,4)

y = -2 x

Find the equation of the line perpendicular to y = 1/2 x - 3 passing through (0,1)

y = -2x + 1

Line A passes through the points (2,-5) and (10,-1) Find the equation of the line perpendicular to A that passes through (4,3)

y = -2x + 11

find the equation of the line perpendicular to y = 1/2 x + 6 through point (3, 5)

y = -2x + 11

find the equation of the straight line passing through the points A(-3, 7) and B(5, -4)

y = -2x + 2

Line A passes through the point (1,5) and (5,7) Find the equation of the line perpendicular to A that passes though (-1,7)

y = -2x + 5

write down the equation of the line perpendicular to y = 1/2 x - 4 which passes through (0, 7)

y = -2x + 7

find the equation of the line with gradient -3 which passes through (4,4)

y = -3x + 16

find the equation of the line parallel to 2y - 3x + 2 = 0 which passes through (0, 4)

y = 1.5x + 4

what is the general line equation for reciprocal graphs

y = 1/ (x-a) + b

write down the equation of the line perpendicular to y = - 3/2 x - 1 which passes through (0, -8)

y = 2/3 x - 8

Find the equation of the line perpendicular to y = 4 - 1/2 x through (2,3)

y = 2x - 1

find the equation of the line parallel to y = 2x + 5 passing through (3,1)

y = 2x - 5

Line A passes through the points (-2,1) and (4,10) find the equation of the line parallel to A that passes through (2,7)

y = 3/2 x + 4

Line A passes through the points (2,1) and (5,10) Find the equation of the line parallel to A that passes through (2,5)

y = 3x - 1

find the equation of the line parallel to y = 3x + 10 through point (4, 6)

y = 3x - 6

Find the equation of the line that is parallel to 2y = 3(2-3x) and passes through the point of intersection of the lines y = x + 8 and y = 3x +4

y = 4.5x + 2.5

find the equation of the line with gradient 4 which passes through (0,2)

y = 4x + 2

find the equation of the straight line parallel to y = 5x - 8 passing through point (0, -1)

y = 5x -1

general equation for a horizontal line

y = a

drawing a quadratic graph

y = ax^2 + bx + c (If y = x^2 - 3 to solve coordinates, draw a table with each individual part - e.g. 1 row with x, x^2, x^2 - 3) (here, x^2 - 3 = the y axis and the x = the x axis)

what is the basic equation for a cubic graph

y = ax^3 + bx^2 + cx + d

what are 2 examples of translations of y = f(x) in trig graphs

y = f(x) + a (translates a squares up) y = f(x) -a (translates a squares down) y = f(x+a) (translates a squares to the left) y = f(x - a) (translates a squared to the right)

equation for a straight line graph

y = mx + c (m = gradient c = the y intercept)

find the equation of the line passing through the points A(2, 4) and B(6, 8)

y = x + 2

how to calculate the gradient

y2 - y1 / x2 - x1 or rise / run