3. Functions and Inverses A. Revision from grade 11 The Straight Line Graph → y = mx + c

m = gradient or Δ𝑦

Δ𝑥 𝑜𝑟 𝑚 =

𝑦2− 𝑦1

𝑥2− 𝑥1

c = is the y-intercept

The quadratic function / parabola will be dealt with in section E.

The hyperbola

→ 𝑦 = 𝑘

𝑥−𝑝+ 𝑞

k = the constant; if k is negative the graph is in the 2nd and 4th

quadrants, and if k is positive the graph falls in the 1st and 3rd

quadrants.

p = is the shift left (x + p) or right ( x – p) of the graph.

x = p is the horizontal axis

q = shift up (+ q) or down (-q) of the graph.

y = q is the vertical asymptote of the graph.

There are two lines of symmetry for the hyperbola →

The gradient is always + 1 or – 1.

To find c (y = mx + c) substitute the point of intersection of the two

asymptotes.

Eg. Given that 𝑓(𝑥) = 3

𝑥−2+ 3 find the equations of the two lines of

symmetry. Gradient / m = ± 1; the point of intersection is (2; 3).

y = - 1x + c or y = 1x + c

3 = -1 (2) + c 3 = 1(2) + c

c = 5 c = 1

y = -x + 5 or y = x + 1

Activity 3.1

1. Below are two graphs, f(x) which is a straight line and g(x) which is a

hyperbola. They intersect at the point A (2;8).

y

A

x

a) Determine the equation of the straight line f(x). b) Determine the equation of the hyperbola given that the x-

asymptote is at 1 and the y-asymptote is at -2. c) Find the other point of intersection of the straight line graph

and the hyperbola. d) Determine the equations for the lines of symmetry for the

hyperbola.

2. Given the equation of the graph f(x) = 2x – 3 and that the second function g(x) is a hyperbola. y

x

a) Give the equations for the two asymptotes. b) Determine the equation of g(x). c) Give the two coordinates where the graphs intersect. d) Determine the two equations of symmetry for the hyperbola. e) For which values of x is f(x) > g(x). f) Give the equation of h(x) if g(x) is shifted two units to the right

and three units down. g) Give the equation of j(x) if f(x) is reflected in the x-axis.

3. The formula f(x) = 𝑎

𝑥+2+ 3 is given for the graph of the hyperbola

where the y-intercept is the point (0; 2). a) Find a. b) Determine the equations of symmetry for f(x). c) Give the x-asymptote of f(x). d) Give the equation of g(x) if g(x) = f(x-3) – 7. e) Give the equation of h(x) which is the graph of f(x) reflected

about the x-axis.

B. Functions Definition: A function occurs when for every value in a given set, there is

only one value in a second set. That is, for every x-value in the domain (set of input values), there is only one corresponding y-value in the range (set of output values).

One-to-one functions: For every x-value, there is only one corresponding y-value. Many-to-one function: There is more than one x-value mapping onto a single y-value. One-to-many function: There is a single x-value mapping onto

more than one y-value (this is not a true function)

To test for a function you can do either a vertical or horizontal line test if you are using a graph (you can do this by placing your ruler vertically on the page and sliding it across the graph → if the ruler touches the graph more than once in the same place the graph is not a function), or you could generate a table of values using the equation. Example: Find out if the function given below is a one-to-one function. That means that no x- or y-value is repeated. Use a table to answer this question. 𝑓(𝑥) = 𝑥2 − 1

x -3 -2 -1 0 1 2 3

y 8 3 0 -1 0 3 8

Therefore this is a many-to-one function because there are two y-values for every x-value

C. Symmetry

You can also do this on your Sharp EL-W535HT by pressing .

Now enter the function by pressing

Your screen will now say X_start and X_step. Enter where you would like to start from for

example -3 and press and the leave the step as 1 so just press again.

Now check whether there are any repeating values and determine whether your graph is a

function.

To find the reflection of a given point (p;q) about the y-axis, make ‘p’ negative → (p; q) reflected about x = 0 (y-axis) → (-p; q)

To find the reflection of a given point (p ; q) about the x-axis, make ‘q’ negative → (p; q) reflected about y = 0 (x-axis) → (p; -q)

To find the reflection of a given point e.g. (p; q) about the line y = x, switch the co-ordinate values around, i.e. (q ; p)

(More of this topic will be covered in paper 2 – Transformations section)

D. Inverse Functions

To find the inverse of a function, we switch the x and y in the equation and then solve for y again. The original domain now becomes the range, and the original range now becomes the domain. The inverse can always be found for a one-to-one function. The graph of the inverse function is reflected about the line, y = x, since the x and y values have been switched. The line y = x is the line of symmetry for a graph and its inverse. The inverse can only be found for a many-to-one function if the values of x are restricted (because the inverse graph becomes a one-to-many graph which is not a function), therefore enabling the resultant function to be the inverse. Example: Find the inverse of 𝑓(𝑥) = 𝑥2 − 1. Switch the x and y values. 𝑓(𝑦) = 𝑦2 − 1 ∴ 𝑥 + 1 = 𝑦2 , Solve for y. ∴ 𝑥 + 1 = 𝑦2

∴ 𝑦 = √𝑥 + 1 = 𝑓−1

The following keys on the EL-W535HT can be used for functions and inverses:

For squaring values

For applying any exponent to a given base

For square rooting a value

For cubing a given value

For entering a fraction

For entering a mixed fraction Activity 3.2 1. Given the following functions, use the Table Mode function on the

EL-W535HT to generate a table, and explain whether these functions are one-to-one or many-to-one/one-to-many.

a) 𝑔(𝑥) = 𝑥 + 3 b) 𝑝(𝑥) = 𝑥3 c) ℎ(𝑥) = 𝑥2 + 𝑥 − 6 d) 𝑘(𝑥) = 𝑥−1

e) 𝑠(𝑥) = √𝑥 2. For each of the following functions, find the inverse. a) 𝑔(𝑥) = 𝑥2 + 7 b) 𝑝(𝑥) = 𝑥3 + 1

c) 𝑠(𝑥) = √𝑥3 3. Write down the domain and range for each of the above equations, and their inverses, so that they are functions.

E. Quadratic Functions

There are three formulae that you can use to find a quadratic function:

y = ax2 +bx + c → this is the standard equation. (In order to find this equation you are usually given two points and you have to substitute and solve for a, b, and c)

y = a(x – p)2 + q → this equation can be used to find the turning point of the graph. p is your x-value turning point and q is your y-value turning point (in order to find this equation you complete the square) – often this form of the equation is given to you and you need to find the other equation (ax2 + bx + c) by multiplying out.

y = a(x – x1)(x – x2) where x1 and x2 are your x-intercepts. To calculate the x-intercepts: remember to make y = 0, and solve for x. If the values cannot be found by factorising the trinomial, use the quadratic

formula: 𝑥 =−𝑏±√𝑏2−4𝑎𝑐

2𝑎

To calculate the y-intercept: make x = 0.

If you are having problems factorising remember that your SHARP EL-W535HT can help you.

Go to table mode. Once there put the equation in as a function.

Start at 1 by pressing and then . Press again so that your step

remains 1. Look in the ANS column for 0 (the x will then be one of your factors).

Remember that when x is positive in the table it is negative in your bracket. For example if

x = 7 and ANS = 0 then your factor is (x – 7).

Once you have this factor you can go back to normal mode. Simply divide the “c” value to find

the other factor. Put the number (or “a” value) that was in front of the x2 in the bracket that

has the second factor.

Remember to think about your signs…

To find the axis of symmetry/turning point:

Axis of symmetry: −𝑏

2𝑎

Turning point (maximum or minimum point): Substitute the value of

−𝑏

2𝑎 into the original equation.

You can also find the turning point by completing the square or changing the formula into the form of y = a(x – p)2 + q where (p;q) is your turning point. The domain is: 𝑥 ∈ (−∞; ∞) The range is either from a minimum value to infinity, or from negative infinity to the maximum value. To find the minimum or maximum find your y-turning point. The shape of your graph is determined by “a” → if a is positive your graph

has a smiley-face shape ( )

If “a” is negative it will have a sad-face shape ( ). If “a” is positive you will have a minimum y-value and if “a” is negative you will have a maximum y-value. If “a” is positive your function’s gradient will first decrease and then increase. If “a” is negative your function’s gradient will first increase and then decrease. Steps for completing the square: - Take out the value of ‘a’ as a common factor, i.e. the coefficient of

𝑥2. - Multiply the coefficient of the x-term by a half, and then square the

answer. - Add this value to create a new trinomial, and then to compensate

for this adding on, subtract the same value from the whole number given in the original equation.

- Write the new equation in the form: 𝑦 = 𝑎(𝑥 − 𝑝)2 + 𝑞 Use this equation to calculate the turning point. To find the x- and y-intercepts, make y = 0 and x = 0 in the original equation. If the

x-values cannot be found through factorisation, use the quadratic formula. Example:

𝑓(𝑥) = 1

2𝑥2 + 𝑥 − 1

Step 1: Take out the common factor of 1

2

𝑓(𝑥) = 1

2𝑥2 + 𝑥 − 1

∴ 𝑓(𝑥) = 1

2(𝑥2 + 2𝑥) − 1

Step 2: Take the coefficient of x (e.g. 2) and substitute it into the axis

of symmetry formula −𝑏

2𝑎 and square this value:

∴ (− 2

2(1))2= 1

Step 3: Add 1 to the bracket and write this as a square, and then

subtract 1 from the constant outside the bracket (remember that you have a common factor of so you need to multiply the 1 you are subtracting by when you take it out of the brackets).

𝑓(𝑥) = 1

2(𝑥2 + 2𝑥 + 1) − 1 −

1

2

∴ 𝑓(𝑥) = 1

2(𝑥 + 1)2 −

3

2

Step 4: The turning point is (−1; −3

2)

Step 5: The x-intercepts: Put the original equation equal to 0.

𝑓(𝑥) = 1

2𝑥2 + 𝑥 − 1 (multiply by 2 to get rid of the fraction).

0 = 𝑥2 + 2𝑥 − 2 This cannot be factorised into 2 brackets,

therefore the quadratic equation must be used.

𝑥 = −𝑏±√𝑏2−4𝑎𝑐

2𝑎

∴ 𝑥 = −1 ± √12−4(

1

2)(−1)

2(1

2)

∴ 𝑥 = −1 + √3 𝑜𝑟 𝑥 = −1 − √3 ∴ 𝑥 = 0.73 𝑜𝑟 𝑥 = −2.73 Step 6: The y-intercept: Put x = 0. Therefore the intercept is (0; -1)

Activity 3.3 For each of the following quadratic functions: (i) Find the turning point

(ii) Find the x and y intercepts

(iii) Find the axis of symmetry

(iv) Write down the domain and range of f(x)

a) 𝑓(𝑥) = 3𝑥2 − 𝑥 − 2 b) 𝑓(𝑥) = 1

4𝑥2 −

5

2𝑥 + 4

c) 𝑓(𝑥) = 2𝑥2 + 7𝑥 + 6 d) 𝑓(𝑥) = 1

2𝑥2 + 2𝑥 − 6

Answers for Activities Activity 3.1 1. a) y = mx + c y = mx + 2 (y-intercept) -4 = m(-2) + 2 -6 = m(-2) m = 3 ∴ 𝑦 = 3𝑥 + 2

b) 𝑦 = 𝑎

𝑥−𝑝+ 𝑞

= 𝑎

2−1− 2

10 = 𝑎

1

a = 10. ∴ 𝑦 = 1

𝑥−1− 2

c) y = 3x + 2 and 𝑦 = 1

𝑥−1− 2

Solve simultaneously

𝑥 + 2 = 1

𝑥−1− 2

3𝑥 + 4 = 1

𝑥−1

3𝑥2 − 3𝑥 + 4𝑥 − 4 = 10 3𝑥2 + 𝑥 − 14 = 0 (x – 2)(3x + 7) = 0

𝑥 = 2 𝑜𝑟 𝑥 = −3

y = 8 ∴ 𝑦 = 3 (−

3) + 2

= - 5 d) y = ± x + c (1; -2) -2 = +1 (1) + c OR -2 = -1 (1) + c c = -3 c = -1

∴ 𝑦 = 𝑥 − 3 𝑜𝑟 𝑦 = −𝑥 − 1

2. a) x = -2 and y = 1

b) 𝑔(𝑥) = 𝑎

𝑥−𝑝+ 𝑞

2 = 𝑎

+2+ 1

1 = 𝑎

2

a = 2 ∴ 𝑔(𝑥) = 2

𝑥+2+ 1

c) y = 2x – 3 and y = 2

𝑥+2+ 1

2𝑥 − 3 =2

𝑥+2+ 1

2𝑥 − 4 = 2

𝑥+2

2𝑥2 − 4𝑥 + 4𝑥 − = 2 2𝑥2 − 10 = 0 𝑥2 − = 0 𝑥2 =

𝑥 = ±√

∴ 𝑦 = +√ 2 − 3 or 𝑦 = −√ 2 − 3 = 1,47 = - 7,47

d) y = ± x + c (-2; 1)

1 = +1 (-2) + c or 1 = -1 (-2) + c c = 3 c = -1

y = x + 3 y = -x -1

e) ℎ(𝑥) = 2

𝑥−2+2+ 1 − 3

= 2

𝑥− 2

f) j(x) = -y = 2x -3

= -2x + 3

3. a) 𝑓(𝑥) = 𝑎

𝑥+2+ 3

2 = 𝑎

+2+ 3

−1 = 𝑎

2

a = -2

b) y = ± x + c (-2; 3) 3 = + 1(-2) + c OR 3 = - 1(-2) + c c = 5 c = 1 y = x + 5 y = -x + 1 c) x = -2

d) 𝑔(𝑥) = −2

𝑥+2−3+ 3 − 7

= −2

𝑥−1− 4

e) ℎ(𝑥) = −𝑦 = −2

𝑥+2+ 3

= 2

𝑥+2− 3

Activity 3.2 1. a)

X -5 -4 -3 -2 -1 0 1 2 3 4 5

Y -2 -1 0 1 2 3 4 5 6 7 8

No x- or y-values are repeated, therefore this is a one-to-one function. b)

X -5 -4 -3 -2 -1 0 1 2 3 4 5

Y -125 -64 -27 -8 -1 0 1 8 27 64 125

No x- or y-values have been repeated, therefore this is a one-to-one function.

c)

X -5 -4 -3 -2 -1 0 1 2 3 4 5

Y 14 6 0 -4 -6 -6 -4 0 6 14 24

Values have been repeated, therefore this is a many-to-one function. d)

X -5 -4 -3 -2 -1 0 1 2 3 4 5

Y -0.2 -0.25 -0.33 -0.5 -1 und 1 0.5 0.33 0.25 0.2

No x- or y-values are repeated, therefore this is a one-to-one function. e)

X -5 -4 -3 -2 -1 0 1 2 3 4 5

Y 0 1 1.4 1.7 2.2 2.3

No x- or y-values are repeated, therefore this is a one-to-one function. 2. a) 𝑔(𝑥) = 𝑥2 + 7 b) 𝑝(𝑥) = 𝑥3 + 1 𝑦 = 𝑥2 + 7 𝑦 = 𝑥3 + 1 𝑥 = 𝑦2 + 7 𝑥 = 𝑦3 + 1 𝑥 − 7 = 𝑦2 𝑥 − 1 = 𝑦3

𝑦 = ± √𝑥 − 7 √𝑥 − 1

= 𝑦

c) 𝑠(𝑥) = √𝑥3

𝑦 = √𝑥3

𝑥 = √𝑦3

𝑥2 = 𝑦3

𝑦 = √𝑥2

𝑜𝑟 𝑦 = 𝑥2

3.

a) 𝑦 = ±√𝑥 − 7 This is not a one-to-one function therefore the domain has to be restricted. Domain: 𝑥 ∈ [7; ∞), Range: [0; ∞)

b) 𝑦 = √𝑥 − 1

This is a one-to-one function therefore there is no need to restrict the domain in order to make it a function. Domain: ∈ (−∞; ∞) , Range: (−∞; ∞)

c) 𝑦 = √𝑥2

This is a one-to-one function therefore there is no need to restrict the domain in order to make it a function. Domain: ∈ (−∞; ∞) , Range: (−∞; ∞). Activity 3.3 a) 𝑓(𝑥) = 3𝑥2 − 𝑥 − 2

(i) TP: x-value → − 𝑏

2𝑎

= −1

2(3)

= 1

6

y-value: substitute x-value into the equation and solve for y:

𝑦 = 3 (1

6)2−

1

6− 2 = −2

1

12

Therefore the turning point is (1

6; −2

1

12)

(ii) X-intercept, make y = 0: 0 = 3𝑥2 − 𝑥 − 2 0 = (3𝑥 + 2)(𝑥 − 1)

Therefore x = 1 or 𝑥 = −2

3

Y-intercept, make x = 0: 𝑓(0) = 3(0)2 − 0 − 2, therefore y = -2

(iii) Axis of symmetry is the line which passes through the turning point to divide the graph in half.

Therefore, 𝑥 = 1

6

(iv) Domain: 𝑥 ∈ (−∞;∞) Range: 𝑦 ∈ [−21

12; ∞)

b) 𝑓(𝑥) = 1

4𝑥2 −

5

2𝑥 + 4

(i) TP: x-value → −(−

5

2)

2(1

4)= . Substitute into the equation to find y.

Therefore 𝑦 = 1

4( )2 −

5

2( ) + 4 = −2

1

4. TP = ( ; −2

1

4)

(ii) X-intercept, make y = 0:

Therefore: 0 = 1

4𝑥2 −

5

2𝑥 + 4

0 = 𝑥2 − 10𝑥 + 16 0 = (𝑥 − 2)(𝑥 − ) x = 2 or x = 8 Y-intercept, make x = 0:

Therefore: 𝑦 = 1

4(0)2 −

5

2(0) + 4 = 4.

(iii) Axis of symmetry is the x-value of the turning point: Therefore x = 5

(iv) Domain: 𝑥 ∈ (−∞; ∞) Range: 𝑦 ∈ [−21

4; ∞)

c) 𝑓(𝑥) = 2𝑥2 + 7𝑥 + 6

(i) TP: x-value −

2(2)= −

4 . Substitute into the equation for the y-value.

𝑦 = 2 (−

4)2+ 7(−

4) + 6 = −

1

8 Therefore: TP = (−

4; −

1

8)

(ii) X-intercepts, make y = 0: Therefore,0 = 2𝑥2 + 7𝑥 + 6 0 = (2𝑥 + 3)(𝑥 + 2)

∴ 𝑥 = −3

2 𝑜𝑟 𝑥 = −2

y – intercept, make x = 0: Therefore, 𝑦 = 2(0)2 + 7(0) + 6 = 6

(iii) The axis of symmetry is 𝑥 = −

4

(iv) Domain: 𝑥 ∈ (−∞;∞) Range: 𝑦 ∈ [−1

8; ∞)

d) 𝑓(𝑥) = 1

2 𝑥2 + 2𝑥 − 6

(i) TP: x-value (−2)

2(1

2)= −2. Substitute into the equation for the

y-value. Therefore 𝑦 = 1

2 (−2)2 + 2(−2) − 6 = − .

The TP is (-2 ; -8) (ii) X-intercept, make y = 0:

Therefore: 0 =1

2 𝑥2 + 2𝑥 − 6 (Multiply by 2 to get rid of

1

2)

0 = 𝑥2 + 4𝑥 − 12 0 = (𝑥 + 6)(𝑥 − 2) ∴ 𝑥 = −6 𝑜𝑟 𝑥 = 2 Y-intercept, make x = 0:

Therefore: 𝑦 = 1

2 (0)2 + 2(0) − 6 = −6

(iii) The axis of symmetry: x = -2 (iv) Domain: 𝑥 ∈ (−∞; ∞) Range: 𝑦 ∈ [− ; ∞)