Week 7 Complete

Question 1

Linear Functions can be written in 2 ways

Answer 1

Standard form: 𝑎𝑥+𝑏𝑦+𝑐=0

Gradient-Intercept form: y=𝑚𝑥+𝑐

Question 2

Write 𝑦=3𝑥+2 in standard form

Answer 2

𝑦=3𝑥+2 𝑦=3𝑥+2(−𝑦) 0=3𝑥−𝑦+2 or 3𝑥−𝑦+2=0

Question 3

The gradient intercept form is so called because

Answer 3

if 𝑦=𝑚𝑥+𝑐 then the gradient of the line is 𝑚 and 𝑐 is the 𝑦 value where the line cuts the y axis.

Question 4

slope is also called the gradient, which is calculated by?

Answer 4

rise/run **(y/x)**

Question 5

linear function? form

Answer 5

y=mx+c 𝑦=(3/2)𝑥+(1)

Question 6

dissect linear function form

Answer 6

y=(3/2)x+1 3/2 = gradient/slope = m 1 = y-int = c

Question 7

y=(3/2)x+1

x intercept (at y = 0)

Answer 7

y=(3/2)x+1

x intercept (at y = 0)

0=(3/2)x+1

• 1 = (3/2)x
• 2/3 = x

Question 8

y=(3/2)x+1

y intercept (at x = 0)

Answer 8

y=(3/2)x+1

y intercept (at x = 0)

y=(3/2)(0)+1 y=1

Question 9

Graph 4𝑥+2𝑦−5=0

using the gradient and y-intercept.

Answer 9

4𝑥+2𝑦−5=0

2𝑦−5 =−4𝑥

2y/2= - 4x/2 +5/2

y= -2x + 2 1/2

m = -2 = - 2/1

(means graph will drop from left to right)

y-intercept = 2 1/2

i.e. (0,2 1/2)

Question 10

Special Linear Equations

There are 2 straight lines that don’t quite fit the form of:

𝑎𝑥+𝑏𝑦+𝑐=0

or

𝑦=𝑚𝑥+𝑐.

Answer 10

1. 𝑥=𝑎
   e. g.

𝑥=3 (or 𝑥−3=0)

This means every point on this line, 𝑥 must be 3.

2. 𝑦=𝑏
   e. g.

𝑦=−2(or 𝑦+2=0)

Likewise, every point on this line 𝑦 must equal −2.

Question 11

the line is positive if it rises from the:

and negative if it rises from the:

Answer 11

positive if Left to Right

negative if Right to Left

Question 12

Answer 12

Question 13

Another method for plotting the graph:
𝑦 = 3/2𝑥 + 1

Answer 13

Question 14

Finding the equation of lines, given one point and the gradient.
Sometimes we know a line passes through a given point and we know the gradient, hence
we can find the equation of this line. This requires the rule:

Answer 14

𝒚 - 𝒚₁ = 𝒎(𝒙 - 𝒙₁)

where (𝑥₁, 𝑦₁) is the point the line passes through and 𝑚 is the gradient

Question 15

Find the equation of the line that passes through (-3,4) with a gradient of 2

Answer 15

(-3,4) with a gradient of 2

(𝑥₁, 𝑦₁) ↔ (-3,4) (i.e. 𝑥₁ = -3, 𝑦₁ = 4)
𝑚 = 2
𝑦 - 𝑦₁ = 𝑚(𝑥 - 𝑥₁)
𝑦 - 4 = 2(𝑥 - (-3))
𝑦 - 4 = 2(𝑥 + 3)
𝑦 - 4 = 2𝑥 + 6
𝒚 = 𝟐𝒙 + 𝟏𝟎
Check: (because (-3,4) lies on this line, when we substitute in 𝑥 = -3 we should get 𝑦 = 4)
𝑦 = 2 × (-3) + 10
𝑦 = -6 + 10
𝑦 = 4
Also 𝑦 = 2𝑥 + 10 has a gradient of 2

Question 16

Find equation of lines through 2 points.
If we know 2 points, we can find gradient, then use 𝑦 - 𝑦₁ = 𝑚(𝑥 - 𝑥₁)
Example: Find the equation of the line that passes though (-3,4) and (2, -1)

Answer 16

Solution: Although its not necessary, we may find it useful to quickly sketch the 2 points on
the Cartesian Plane. The gradient of the line joining 2 points (𝑥₁, 𝑦₁) and (𝑥₂, 𝑦₂)

is

m=y₂-y₁/x₂-x₁

(rise/run)

Question 17

Parallel lines have the same gradient.

Answer 17

𝑦 = 3𝑥 - 2 𝑦 = 3𝑥 + 1 6𝑥 - 2𝑦 + 5 = 0
all have a gradient of 3 and when sketched will be parallel to each other

Question 18

Find the equation of the line passing though (-4, -2) which is parallel to
3𝑥 - 5𝑦 + 8 = 0

Answer 18

Solution: We know parallel lines have the same gradient, so we need to find the gradient of
the given line.

Question 19

𝑦 = 𝑚₁𝑥 + 𝑐₁

𝑦 = 𝑚₂𝑥 + 𝑐₂

product of gradient? perpendicular

Answer 19

𝑚₁ × 𝑚₂ = -1

because they are perpendicular, always -1

Question 20

Find equation of line through (3,4) that is perpendicular to the line
𝑦 =2/5 𝑥 - 1

Answer 20

Question 21

If 2 lines are not parallel to each other

Answer 21

then they will intersect each other at one point (𝑥, 𝑦)
only.
Point of Intersection or the Simultaneous
Solution.
There are 2 methods of finding this
unique solution, Substitution and
Elimination.

Question 22

Solve

3𝑥 + 2𝑦 - 6 = 0
and

2𝑥 - 𝑦 + 10 = 0

by substitution

Answer 22

The substitution method entails rearranging one of the equations to get either 𝑥 or 𝑦 by itself
and then substituting that value into the other equation.

Solution:

3𝑥 + 2𝑦 - 6 = 0 –(1)
2𝑥 - 𝑦 + 10 = 0 –(2)
Number both equations, and make sure
both are in the form 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0

Upon inspection we can see that in equation (2) it can be easily rearranged to make 𝑦 the
subject.

(2)→ 2𝑥 + 10 = 𝑦 –(2a)

Now substitute for 𝑦 in equation (1)

(1)→ 3𝑥 + 2(2𝑥 + 10) - 6 = 0
3𝑥 + 4𝑥 + 20 - 6 = 0

7𝑥 + 14 = 0
7𝑥 = -14
𝑥 = -2

Now substitute for 𝑥 = -2 in (2a)
2𝑥 - 2 + 10 = 𝑦

6 = 𝑦

Answer:

𝒙 = -𝟐
𝒚 = **6**

Question 23

The Elimination method involves

Answer 23

adding or subtracting the equations

Question 24

Solve

2𝑥 + 7𝑦 = 16

and

3𝑥 - 6𝑦 = 2

by elimination

Answer 24

write each equation in standard form

2𝑥 + 7𝑦 - 16 = 0 –(1)
3𝑥 - 6𝑦 - 2 = 0 –(2)

We need to make the co-efficients of either 𝑥 or 𝑦 the same

(1) × 3 6𝑥 + 21𝑦 - 48 = 0 –(1a)
(2) × 2 6𝑥 - 12𝑦 - 4 = 0 –(2a)
(1a) – (2a) 33𝑦 - 44 = 0

33y = 44

y = 44/33 = 4/3

Sub for 𝑦 = 4/3
in either (1) or (2) to find 𝑥

in (1) 2x+7*4/3 -16=0

2x+28/3 - 48/3 = 0

2x=20/3

x=10/3

Answer:

x = 10/3

y = 4/3

Question 25

Solutions to Quadratic Equations Quadratic functions can be written 𝑦=𝑎𝑥²+𝑏𝑥+𝑐 y as a function of x

Answer 25

y = ax²+bx+c [If 𝑎 = 0, then 𝑦 = 𝑏𝑥 + 𝑐 and this is a linear equation or function, so not quadratic.] Now 𝑦 is a function of 𝑥, i.e. 𝑦 = 𝑓(𝑥) Hence a quadratic can also be written 𝑓(𝑥) = 𝑎𝑥² + 𝑏𝑥 + 𝑐

Question 26

Graphing Quadratics: Let’s graph 𝑦=𝑥²+2𝑥−3

Answer 26

This is the shape of the parabola. All quadratics take this shape. It is symmetrical (a mirror image) on either side of 𝑥 = -1 Note that the y-intercept is -3. Recall from work on linear functions, that with all functions 𝑥 intercept is when 𝑦 = 0 𝑦 intercept is when 𝑥 = 0 𝑦 = 𝑥² + 2𝑥 - 3 𝑦 int. (𝑥 = 0) 𝑦 = 0² + 2 × 0 - 3 So the 𝑐 value still gives the 𝑦 intercept The 𝑥 intercept is when 𝑦 = 0 0 = (𝑥 + 3)(𝑥 - 1) Either 𝑥 = -3 or 𝑥 = 1 and these are the 2 points where the graph intersects the 𝑥 axis. The line of symmetry is 𝑥 = - 𝑏/2𝑎 for 𝑦 = 𝑎𝑥² + 𝑏𝑥 + 𝑐 In 𝑦 = 𝑥2 + 2𝑥 - 3, 𝑎 = 1, 𝑏 = 2, 𝑐 = -3 𝑥 = -2/2×1 = -2/2 = -1 We can also find the Turning Point of the parabola, i.e. then point where the function has the greatest or the least value. For this example, it will be the least or minimum value. Because 𝑥 = -1 is the line of symmetry, then the least value of the function 𝑦 = 𝑥² + 2𝑥 - 3 will occur at 𝑥 = -1 Substitution 𝑥 = -1 into this equation gives 𝑦 = (-1)² + 2 × (-1) - 3 = 1 - 2 - 3 = -4 So the Turning point is (-1, -4), as can be seen in the previous graph and the minimum value of the function is -4

Question 27

Sketch the graph 𝑦=−9+6𝑥−𝑥² showing all 𝑥 and 𝑦 intercepts. As well state the Domain and Range of this function

Answer 27

Line of symmetry 𝑥=−𝑏/2𝑎 𝑥=−6/2×−1=3 When 𝑥=3, AND 𝑦=−9+6×3−3² = −9+18−9 = 0 So this quadratic only cuts the 𝑥 axis once at 𝑥=3. Note that the parabola is upside down. When “𝑎” or the coefficient of 𝑥² is negative, the parabola is always upside down Domain of 𝑦=−9+6𝑥−𝑥² All possible Real 𝑥 values −∞\<𝑥\<∞ Range: The greatest value 𝑦 can take is 0, which occurs at the Turning Point. −∞\<𝑦≤0

Question 28

The Discriminant

Answer 28

𝑏² - 4𝑎𝑐 in the Quadratic Formula is called the Discriminant. By calculating 𝑏² - 4𝑎𝑐, we can determine how many times the quadratic function cuts the 𝑥 axis.

Question 29

``` The number (𝑁) of motorists detected speeding on the M1 𝑑 days (1 ≤ 𝑑 ≤ 10) after the installation of a speed camera can be modelled by the equation: 𝑁(𝑑) = 𝑑<sup>2</sup>/2 - 3𝑑 + 75 ``` Find a) When least speeding violations occur b) When most speeding violations occur

Answer 29

𝑎 = 1/2, 𝑏 = -3, 𝑐 = 75 Line of symmetry 𝑑 = - 𝑏/2𝑎 𝑑 = - -3/2×1/2 = 3 𝑁(3) = 3²/2 - 3 × 3 + 75 𝑁(3) = 4 1/2 - 9 + 75 = 70 1/2 So (3, 70 1/2) is the Turning Point. Sub in the end values of 𝑑, i.e 𝑑 = 1, 𝑑 = 10 𝑑 = 1 𝑁(1) = 1²/2 - 3 × 1 + 75 = 1/2 - 3 + 75 = 72 1/2 𝑑 = 10 𝑁(10) = 10²/2 - 3 × 10 + 75 = 50 - 30 + 75 = 95 * *a) Least violations on day 3 (70 1/2) b) Most violations on day 10 (95)**

Question 30

A ball is thrown vertically upwards such that its height (metres) above the ground is: ℎ(𝑡) = 40𝑡 - 5𝑡² where 𝑡 is the number of seconds after it is thrown Find: a) The maximum height reached b) The times when it is 15m high c) The time taken to hit the ground

Answer 30

**a)** Axis of symmetry 𝑡 = - 𝑏/2𝑎 = -40/2×-5 = -40/-10 = 4 At 𝑡 = 4, Turning Point occurs. Sub in ℎ(𝑡) ``` ℎ(4) = 40 × 4 - 5 × 4<sup>2</sup> ℎ(4) = 160 - 80 = 80m ``` **b)** ℎ = 15 so 15 = 40𝑡 - 5𝑡² 5𝑡² - 40𝑡 + 15 = 0 ÷ 5 𝑡² - 8𝑡 + 3 = 0 𝑎 = 1, 𝑏 = -8, 𝑐 = 3 𝑡 = -𝑏±√𝑏²-4𝑎𝑐/2𝑎 𝑡 = 8±√(-8)²-4×1×3/2×1 𝑡 = 8±√64-12/2 = 8+√52/2 or 8-√52/2 𝑡 = 7.61secs or 0.39secs **c)** ℎ = 0 0 = 40𝑡 - 5𝑡² 0 = 5𝑡(8 - 𝑡) 𝑡 = 0 or 𝑡 = 8 𝑡 = 0 corresponds to when the ball is thrown So 𝑡 = 8 or 8 seconds before it hits the ground

Question 31

Linear and Quadratic Functions are examples of functions called?

Answer 31

**_polynomials._** **Linear**: 𝑓(𝑥) = 𝑚𝑥 + 𝑐 𝑚 ≠ 0 **Quadratic**: 𝑓(𝑥) = 𝑎𝑥² + 𝑏𝑥 + 𝑐 𝑎 ≠ 0 **Cubic**: 𝑓(𝑥) = 𝑎𝑥³ + 𝑏𝑥² + 𝑐𝑥 + 𝑑 𝑎 ≠ 0 **Quartic**: 𝑓(𝑥) = 𝑎𝑥⁴ + 𝑏𝑥³ + 𝑐𝑥² + 𝑑𝑥 + 𝑒 𝑎 ≠ 0

Question 32

**Graphical Transformation** It is possible to change the shape and position of a function by?

Answer 32

changing the co-efficients of 𝑥 and the constant term

Question 33

**Linear Transformation** Consider graphs of the form 𝑦 = 𝑚𝑥 + 𝑐 (𝑐 = 0). They will all pass through the origin (0,0)

Answer 33

𝑚 = 1 𝑦 = 𝑥 will pass through origin at 45˚ 𝑚 = 2 𝑦 = 2𝑥 Is twice as steep 𝑚 = 1/2 𝑦 = 1/2x Is half as steep and so on 𝑚 = -1 𝑦 = -𝑥 will pass through origin from negative direction (i.e. top left at 45˚ 𝑚 = -2 𝑦 = -2𝑥 Is twice as steep as 𝑦 = -𝑥 𝑚 = -1/2 𝑦 = -1/2𝑥 Is half as steep

Question 34

The value of 𝑐 moves the line vertically. Start with 𝑦 = 𝑥

Answer 34

𝑦 = 𝑥 + 3 is parallel to 𝑦 = 𝑥 and moved units 3 up on the y axis 𝑦 = 𝑥 - 2 moves 𝑦 = 𝑥 2 units down

Question 35

Sketch 𝑦 = -2𝑥 + 3 showing all transformations from 𝑦 = -𝑥

Answer 35

1. 𝑦 = -2𝑥 2. 3 units up 𝑦 = -2𝑥 + 3

Question 36

**Transformations of Quadratics** Let’s start with a simple quadratic, 𝑦 = 𝑥²

Answer 36

𝑦 = 2𝑥² is twice as steep 𝑦 = 1/2\*\*𝑥² is half as steep Now 𝑦 = -𝑥² and 𝑦 = -2𝑥² and 𝑦 = -1/2\*\*𝑥²

Question 37

We can move these quadratic graphs vertically

Answer 37

by adding and subtracting values Adding 2 to 𝑦 = 2𝑥² moves 𝑦 = 2𝑥² 2 units up the y axis 𝑦 = 2𝑥² + 2 Subtracting 3 from 𝑦 = 2𝑥² gives 𝑦 = 2𝑥² - 3 and the graph moves 3 units down the y axis

Question 38

Show all transformations, starting with 𝑦 = 𝑥², to sketch 𝑦 = - 1/3\*\*𝑥² - 2

Answer 38

1. 𝑦 = 𝑥² 2. 𝑦 = 1/3\*\*𝑥² 3. 𝑦 = - 1/3\*\*𝑥² 4. 𝑦 = - 1/3\*\*𝑥² - 2

Question 39

**Horizontal Transformations** To move 𝑦 = 𝑥² a) 2 units to right? b) 3 units to left

Answer 39

a) y = (x - 2)² b) 𝑦 = (𝑥 + 3)²

Question 40

The graph 𝑦 = (𝑥 - 1)² + 4

Answer 40

1. 𝑦 = 𝑥² 2. 𝑦 = (𝑥 - 1)² 3. 𝑦 = (𝑥 - 1)² + 4

Question 41

Write in standard from y = 3/5x + 2

Answer 41

y = 3/5x + 2 3/5x - y + 2 = 0 3x - 5y + 10 = 0

Question 42

Write in gradient-intercept from 3𝑥 - 2𝑦 + 5 = 0

Answer 42

3𝑥 - 2𝑦 + 5 = 0 2y = 3x + 5 y = 3/2x + 5/2

Question 43

Write in gradient-intercept from 𝑥 + 3𝑦 - 8 = 0

Answer 43

𝑥 + 3𝑦 - 8 = 0 3y = -x + 8 y = -1/3\*\*x + 8/3

Question 44

state the gradient and y intercept 2𝑥 - 3𝑦 - 1 = 0

Answer 44

2x - 3y - 1 = 0 3y = 2x - 1 y = 2/3\*\*x - 1/3

Question 45

state the gradient and y intercept 𝑦 + 3𝑥 = 4

Answer 45

y + 3x = 4 y = -3x + 4

Question 46

state the gradient and y intercept y = 6

Answer 46

``` y = 6 y = 0x + 6 ```

Question 47

Sketch each of these functions showing the 𝑥 and 𝑦 intercepts a) 𝑦 = 4𝑥 - 8

Answer 47

x-intercept is 2, y-intercept is -8, gradient is 4

Question 48

Sketch each of these functions showing the 𝑥 and 𝑦 intercepts b) 3𝑥 - 4𝑦 - 2 = 0

Answer 48

3𝑥 - 4𝑦 - 2 = 0 y = 3x/4 - 1/2 x int 2/3, y int -1/2, gradient 3/4

Question 49

Sketch each of these functions showing the 𝑥 and 𝑦 intercepts 2x + y + 1 = 0

Answer 49

2x + y + 1 = 0 y = -2x - 1 x-intercept is -1/2 y-intercept is -1 gradient is -2

Question 50

Sketch each of these functions showing the 𝑥 and 𝑦 intercepts 𝑦 = -3𝑥 + 2

Answer 50

x-int 2/3 y-int 2 gradient -3

Question 51

Find the equation of the line that a) passes through (-1,4) with gradient 3

Answer 51

y - y₁ = m(x - x₁) y - 4 = 3(x + 1) y - 4 = 3x + 3 y = 3x + 7

Question 52

Find the equation of the line that b) passes through (2, -3) with gradient -1

Answer 52

y - y₁ = m(x - x₁) y + 3 = -1(x - 2) y + 3 = -x + 2 y = -x - 1

Question 53

Find the equation of the line that c) passes through (-3,1) and (4, -5)

Answer 53

We will start by finding the gradient, m, of the line. m = y2-y1/x2-x1 m = -5-1/4+3 m = -6/7 Now we can find the equation of the line that passes through the two points. y - y1 = m(x - x1) y -1 = -6/7(x + 3) y -1 = -6/7\*\*x - 18/7 y = -6/7\*\*x - 11/7

Question 54

Find the equation of the line that d) passes through (-4,0) and (0,3)

Answer 54

start by finding the gradient of the line m = y₂-y₁/x₂-x₁ = 3-0/0+4 = 3/4 Now we can find the equation of the line that passes through the two points. y - y₁ = m(x - x₁) y - 0 = 3/4(x + 4) y = 3/4\*\*x + 3

Question 55

Find the equation of the line that e) passes through (1,8) and parallel to 𝑦 = 2/3\*\*𝑥 + 4

Answer 55

If the line is parallel to y = 2/3\*\*x + 4, then the gradient of the line will be 2/3 y - y1 = m(x - x1) y - 8 = 2/3\*\*(x - 1) y - 8 = 2/3\*\*x - 2/3 y = 2/3\*\*x + 22/3

Question 56

Find the equation of the line that f) passes through (-2, -4) and parallel to 5𝑥 - 𝑦 + 1 = 0

Answer 56

We will start by finding the gradient of 5x - y + 1 = 0 5x - y + 1 = 0 y = 5x + 1 Therefore, the gradient is 5. y - y1 = m(x - x1) y + 4 = 5(x +2) y +4 = 5x + 10 y = 5x + 6

Question 57

g) passes through (5,1) and perpendicular to 3𝑥 + 2𝑦 = 0

Answer 57

We will start by finding the gradient of 3x + 2y = 0 3x + 2y = 0 2y = -3x y = -3/2\*\*x Therefore, as the line we are trying to find it perpendicular to 3x+2y = 0, the gradient of the line we are trying to find must be 2/3. (Remember the gradients of perpendicular lines multiply to give -1). y-y1=m(x-x1) y - 1 = 2/3\*\*(x-5) y - 1 = 2/3\*\*x - 10/3 y = 2/3\*\*x - 7/3

Question 58

Solve these simultaneous equations (then check your answers) a) -2𝑥 + 7𝑦 = 8 𝑦 + 1 = 3𝑥

Answer 58

-2x + 7y = 8 (1) y + 1 = 3x (2) Rearranging the second equation gives -2x +7y = 8 (1) 3x -y = 1 (2) To eliminate the y term, we will calculate (1) + 7 × (2). 19x = 15 x = 15/19 To find y, we will substitute x = 15/19 into Equation (2). (It doesn’t matter which equation is used to solve for y. Equation (2) was chosen as it appeared to be the easiest to use to solve for y). y + 1 = 3 × 15/19 y + 1 = 45/19 y = 26/19 Thus, x = 15/19 y = 26/19

Question 59

Solve these simultaneous equations (then check your answers) b) 2𝑥 - 𝑦 - 11 = 0 4𝑥 + 𝑦 - 8 = 0

Answer 59

2x - y - 11 = 0 (1) 4x + y - 8 = 0 (2) Simply adding Equations (1) and (2) will eliminate the y term 6x - 19 = 0 6x = 19 x = 19/6 to find y we will substitute x = 19/6 into Equation (1) 2 \* 19/6 - y - 11 = 0 -y - 14/3 = 0 y = -14/3 Thus, x = 19/6 y = -14/3

Question 60

Solve these simultaneous equations (then check your answers) c) -3𝑥 + 𝑦 - 8 = 0 4𝑥 + 2𝑦 - 21 = 0

Answer 60

-3x + y - 8 = 0 (1) 4x + 2y - 21 = 0 (2) To eliminate the y term, we will calculate 2 × (1) - (2). -10x + 5 = 0 10x = 5 x = 1/2 To find y, we will substitute x = 1/2 into Equation (1) -3 × 1/2 + y - 8 = 0 y - 19/2 = 0 y =19/2 Therefore, ``` x = 1/2 y = 19/2 ```

Question 61

Solve these simultaneous equations (then check your answers) d) 4𝑥 - 7𝑦 = 21 6 = -2𝑥 + 𝑦

Answer 61

4x - 7y = 21 (1) -2x + y = 6 (2) To eliminate the x term, we will calculate (1) + 2 × (2) -5y = 33 y = -33/5 To find x, we will substitute y = -33/5 into Equation (1). 4x - 7 \* -33/5 = 21 4x + 231/5 = 21 4x = -126/5 x = -63/10 thus x = -63/10 y = -33/5

Question 62

a) fill in the table b) Plot the 9 ordered pairs (𝑥, 𝑦) on the Cartesian plane c) Draw a smooth curve through all 9 points d) Calculate and draw in the line of symmetry e) Label all 𝑥 and 𝑦 intercepts f) Calculate and label the co-ordinates of the Turing Point. g) State the Domain and Range of 𝑓(𝑥) = -𝑥² + 3𝑥 + 4

Answer 62

y = -x² + 3x + 4 x -4 -3 -2 -1 0 1 2 3 4 y -24 -14 -6 0 4 6 6 4 0 (b) See graph below. (c) See graph (d) The line of symmetry is at x = 1.5. It is shown on the graph using a red, dashed line. (e) See graph (f) There is a turning point, a maximum, at (1.5, 6.25). (g) Domain: All real values of x, Range: y ≤ 6.25 | (a)

Question 63

Use the Discriminant to determine the number of intercepts on the 𝑥 axis a) 𝑦 = 3𝑥² + 4𝑥 - 1

Answer 63

*To find the discriminant, D, of the quadratic y = ax² + bx + c, calculate D = b² - 4ac.* If the discriminant is **positive** (D\>0), the quadratic has **two x-intercepts.** If the discriminant is **zero** (D = 0), the quadratic only has **one x-intercept.** If the discriminant is **negative** (D\<0), the quadratic has **no x-intercepts.** ``` y = 3x<sup>2</sup> + 4x - 1 D = 4<sup>2</sup> - 4 × 3 × -1 = 28 ``` As the discriminant is positive, the quadratic has two x-intercepts.

Question 64

Use the Discriminant to determine the number of intercepts on the 𝑥 axis b) 𝑦 = 𝑥² + 4𝑥 + 4

Answer 64

``` y = x<sup>2</sup> + 4x + 4 D = 4<sup>2</sup> - 4 × 1 × 4 = 0 ``` As the discriminant is zero, the quadratic has one x-intercept.

Question 65

Use the Discriminant to determine the number of intercepts on the 𝑥 axis c) 𝑦 = -𝑥² - 10𝑥 + 2

Answer 65

y = -x² - 10x + 2 D = (-10)² - 4 × -1 × 2 = 108 As the discriminant is positive, the quadratic has two x-intercepts.

Question 66

Use the Discriminant to determine the number of intercepts on the 𝑥 axis d) 𝑦 = -2𝑥² + 𝑥 - 1

Answer 66

y = -2x² + x - 1 D = 1² - 4 × -2 × -1 = -7 As the discriminant is negative, the quadratic has no x-intercepts.

Question 67

Show all transformations in a sketch a) Start with 𝑦 = -𝑥 Finish with 𝑦 = -2𝑥 + 3

Answer 67

* y = -x is drawn in blue * y = -2x is drawn in red * y = -2x + 3 is drawn in black To draw a graph of y = -2x+3, it may be easier to just start with a graph of y = -2x.

Question 68

Show all transformations in a sketch b) Start with 𝑦 = 𝑥² Finish with 𝑦 = 2𝑥² - 4

Answer 68

* y = x² is drawn in blue * y = 2x² is drawn in red • y = 2x² - 4 is drawn in black To draw a graph of y = 2x² -4, it may be easier to just start with a graph of y = 2x²

Question 69

Show all transformations in a sketch c) Start with 𝑦 = 𝑥² Finish with 𝑦 = -1/4\*\*𝑥² + 2

Answer 69

* y = x² is drawn in blue * y = -1/4\*\*x² is drawn in red * y = -1/4\*\*x² + 2 is drawn in black

Question 70

Show all transformations in a sketch d) Start with 𝑦 = 𝑥² Finish with 𝑦 = (𝑥 - 3)² + 1

Answer 70

* y = x² is drawn in blue * y = (x - 3)² is drawn in red * y = (x - 3)² + 1 is drawn in black

Question 71

The volume of water in a tank, 𝑉(𝑡) m³, over a 10 month period is given by the function 𝑉(𝑡) = 2𝑡² - 16𝑡 + 40 where 𝑡 is in months and 𝑡 ∈ [0,10] Find a) The volume when 𝑡 = 0

Answer 71

V (0) = 2 × 0² - 16 × 0 + 40 = 40 Initially (t = 0) there is 40 m³ of water in the tank

Question 72

The volume of water in a tank, 𝑉(𝑡) m³, over a 10 month period is given by the function 𝑉(𝑡) = 2𝑡² - 16𝑡 + 40 where 𝑡 is in months and 𝑡 ∈ [0,10] b) The volume when 𝑡 = 10

Answer 72

V (10) = 2 × 10² - 16 × 10 + 40 = 80 After 10 months (t = 10) there is 80 m³ of water in the tank.

Question 73

The volume of water in a tank, 𝑉(𝑡) m³, over a 10 month period is given by the function 𝑉(𝑡) = 2𝑡² - 16𝑡 + 40 where 𝑡 is in months and 𝑡 ∈ [0,10] c) The minimum volume over the 10 months

Answer 73

To find the minimum volume in the tank over the first 10 months, we will start by drawing a graph of V (t). From the graph, it can be seen that the minimum volume in the tank occurs at t = 4. V (4) = 2 × 4² - 16 × 4 + 40 = 8 Alternatively, if we didn’t want to draw a graph, the x value of the turning point is given by x = -b/2a = 16/2\*2 = 4 Thus the minimum volume in the tank over the 10 month period is 8 m³

Question 74

The volume of water in a tank, 𝑉(𝑡) m³, over a 10 month period is given by the function 𝑉(𝑡) = 2𝑡² - 16𝑡 + 40 where 𝑡 is in months and 𝑡 ∈ [0,10] d) The maximum volume over the 10 months

Answer 74

To find the maximum volume in the tank over the 10 month period, we need to test the end points, t = 0 and t = 10. The maximum volume will not occur at the turning point as it was a minimum. V (0) = 40 and V (10) = 80 (found in part (a)). Therefore the maximum volume of water in the tank over the 10 month period is 80 m³

Question 75

The parabolic Arch of a bridge spans a river 120m wide. The highest point of the arch is 30m above the water. Calculate a mathematical model ℎ(𝑑) that describes the height of the arch above the water at any point 𝑑 metres across the river.

Answer 75

The graph is a little hard to see, but we have the following information about the function h(d) where h is the height of the arch, in metres, above the water and d is the distance, in metres, across the river. ``` h(0) = 0 h(60) = 30 h(120) = 0 ``` ``` We have also been told that h(d) is a parabolic function (a quadratic). Let h(d) = ad<sup>2</sup> + bd + c ``` where a, b and c are constants that we need to determine. Using h(0) = 0 gives h(0) = a × 0² + b × 0 + c = 0 c = 0 Using h(120) = 0 gives h(120) = a × 120² + b × 120 = 0 b × 120 = -a × 120² b = -120a ``` Using h(60) = 30 and b = -120a gives h(60) = a × 60<sup>2</sup> + b × 60 = 30 3600a + 60 × -120a = 30 3600a - 7200a = 30 -3600a = 30 a = -1/120 ``` Finally, using b = -120a and that a = -1/120 gives b = -120 \* -1/120 = 1 Therefore, putting all the information together, the function h(d) is given by h(d) = ad² + bd + c = -1/120 × d² + 1 × d + 0 = -d²/120 + d