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LemonmathSelf-study sheet
Grade 9 · Lesson 6.1

The function y = ax² (a ≠ 0)

Name: Date: Time spent: min

Goals

Remember

Exercises

Exercise 1

Consider the function y=3x2y=3x^2. Find the value of yy when x=−2x=-2, x=0x=0, x=13x=\dfrac13 and x=3x=\sqrt3.

Exercise 2

Make a table of values and draw the graph of y=−12x2y=-\dfrac12x^2 for x∈{−4,−2,0,2,4}x\in\{-4, -2, 0, 2, 4\}. Describe the position of the graph.

Exercise 3

Which of the points A(1,2)A(1, 2), B(−2,8)B(-2, 8), C(3,18)C(3, 18), D(−1,−2)D(-1, -2) lie on the graph of y=2x2y=2x^2?

Exercise 4

The graph of y=ax2y=ax^2 passes through the point M(−3,6)M(-3, 6).

a) Find aa.

b) Find the points on the graph whose yy-coordinate is 2424.

Exercise 5

An object is dropped from a height of 125125 m. The distance it has fallen after tt seconds is s=5t2s=5t^2 (m).

a) After 33 seconds, how far above the ground is the object?

b) How long does it take to reach the ground?

Exercise 6

In grade 9 science (the lesson on kinetic and potential energy), the kinetic energy of an object of mass mm (kg) moving at speed vv (m/s) is W=12mv2W=\dfrac12mv^2 (J). Suppose a ball has mass 0.40.4 kg; then W=0.2v2W=0.2v^2.

a) Find the kinetic energy of the ball when v=10v=10 m/s.

b) The ball has kinetic energy 4545 J. What is its speed?

Exercise 7

Consider the function y=3x2y=3x^2. (Revise square-root expressions.)

a) Find the xx-coordinates of the points on the graph whose yy-coordinate is 66.

b) Find the value of yy when x=1−2x=1-\sqrt2.

Exercise 8

(2.0 marks) The graph (P)(P) of the function y=ax2y=ax^2 passes through the point A(2,2)A(2, 2).

a) Find aa.

b) Draw (P)(P).

c) Find the points on (P)(P) whose yy-coordinate is twice their xx-coordinate.

Exercise 9 Challenge

Suppose an entrance arch has the shape of a parabola. It is 88 m wide at the base and its top is 66 m above the ground. Choose axes with the origin OO at the top of the arch and the yy-axis pointing upwards; then the arch is part of the graph of y=ax2y=ax^2. A lorry 33 m wide and 55 m high drives through the middle of the arch. Can the lorry get through?

Mini test

15 minutes, without looking back at the lesson. Part I: circle one letter; Part II: write T or F for each statement; Part III: write the answer in the box.

Part I. Multiple choice

Question 1. Which of the following functions has the form y=ax2y=ax^2 with a≠0a\neq0?
A. y=3xy=3x.B. y=−12x2y=-\dfrac12x^2.C. y=x2+2y=x^2+2.D. y=0x2y=0x^2.
Question 2. Consider the function y=−2x2y=-2x^2. The value of yy when x=−3x=-3 is
A. 1818.B. −18-18.C. 3636.D. 66.
Question 3. The graph of y=ax2y=ax^2 passes through the point (−2,8)(-2, 8). The value of aa is
A. 22.B. −2-2.C. 44.D. −4-4.
Question 4. Which of the following points does not lie on the graph of y=−x2y=-x^2?
A. (1,−1)(1, -1).B. (−2,−4)(-2, -4).C. (3,9)(3, 9).D. (0,0)(0, 0).

Part II. True or false

Question 1. The function y=13x2y=\dfrac13x^2 has graph (P)(P). Decide whether each statement is true or false.
a) O(0,0)O(0, 0) is the lowest point of (P)(P).T / F
b) (P)(P) passes through the point (3,3)(3, 3).T / F
c) (P)(P) passes through the point (−3,−3)(-3, -3).T / F
d) The points of (P)(P) with yy-coordinate 1212 have xx-coordinates 66 and −6-6. T / F

Part III. Short answer

Question 1. An object falls from a height of 8080 m, and the distance fallen after tt seconds is s=5t2s=5t^2 (m). After how many seconds does it reach the ground?
Question 2. The graph of y=ax2y=ax^2 passes through the point A(4,4)A(4, 4). Find aa (give your answer as a decimal).

Score: ........ / 10

Hints (only when stuck)

Exercise 1.

What you need: substitute the value of xx into the formula, square first, then multiply by 33.

Steps:

  1. Work out x2x^2 for each value.
  2. Multiply the result by 33.
Exercise 2.

What you need: if a<0a<0, the parabola lies below the xx-axis and OO is its highest point.

Steps:

  1. Work out yy at each value of xx.
  2. Plot the points and join them with a smooth curve.
Exercise 3.

What you need: M(x0,y0)M(x_0, y_0) lies on the graph exactly when y0=2x02y_0=2x_0^2.

Steps:

  1. Work out 2x022x_0^2 for each point and compare it with y0y_0.
Exercise 4.

What you need: substitute the coordinates into y=ax2y=ax^2; the equation x2=kx^2=k (k>0k>0) has two solutions ±k\pm\sqrt k.

Steps:

  1. Solve 9a=69a=6.
  2. Solve ax2=24ax^2=24.
Exercise 5.

What you need: the height above the ground is 125−s125-s; the object reaches the ground when s=125s=125.

Steps:

  1. Work out ss when t=3t=3.
  2. Solve 5t2=1255t^2=125 with t>0t>0.
Exercise 6.

What you need: W=0.2v2W=0.2v^2 is a function of the form y=ax2y=ax^2 with a=0.2a=0.2; the speed v>0v>0.

Steps:

  1. Substitute v=10v=10.
  2. Solve 0.2v2=450.2v^2=45.
Exercise 7.

What you need: x2=2⇔x=±2x^2=2\Leftrightarrow x=\pm\sqrt2; the identity (a−b)2=a2−2ab+b2(a-b)^2=a^2-2ab+b^2.

Steps:

  1. Solve 3x2=63x^2=6.
  2. Expand (1−2)2(1-\sqrt2)^2, then multiply by 33.
Exercise 8.

What you need: points on a graph; a product equation A⋅B=0A\cdot B=0 (lesson equations reducible to linear equations).

Steps:

  1. Solve 4a=24a=2.
  2. Make a table of values with x∈{−4,−2,0,2,4}x\in\{-4, -2, 0, 2, 4\}.
  3. The required points have the form (x,2x)(x, 2x): solve 12x2=2x\dfrac12x^2=2x.
Exercise 9.

What you need: the feet of the arch are the points (−4,−6)(-4, -6) and (4,−6)(4, -6).

Steps:

  1. Find aa from the point (4,−6)(4, -6).
  2. The sides of the lorry are 1.51.5 m from the centre line: find yy at x=1.5x=1.5.
  3. The height of the arch there is 6+y6+y; compare it with 55.

Self-check

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