The function y = ax² (a ≠ 0)

Between about 1602 and 1604, in Padua (Italy), Galileo Galilei rolled balls down sloping channels to slow down falling motion so that he could measure it. He found a surprising law: the distance travelled is not proportional to the time, but to the square of the time (MacTutor, University of St Andrews). In grade 10 physics, the law of free fall is written as $s=\dfrac12gt^2$ with $g\approx9.8$ m/s²; rounding to $g\approx10$ gives $s\approx5t^2$ (metres).

Suppose you drop a marble from the roof of a building $45$ m high (about $15$ storeys) and ignore air resistance. How long does it take the marble to reach the ground? And why does it fall so much further in the third second than in the first?

The function $y=ax^2$ ( $a\neq0$ )

Explore: distance fallen over time

We work out $s=5t^2$ at a few times:

$t$ (seconds)	$0$	$1$	$2$	$3$
$s$ (metres)	$0$	$5$	$20$	$45$

When the time doubles (from $1$ to $2$ seconds), the distance becomes four times as large (from $5$ to $20$ m); three times the time gives nine times the distance. In each second the marble falls $5$ m, then $15$ m, then $25$ m: it keeps speeding up. Each value of $t$ gives exactly one value of $s$ , so $s$ is a function of $t$ .

Going deeper. For the linear function $y=ax$ (grade 8), doubling $x$ doubles $y$ . For $y=ax^2$ , multiplying $x$ by $k$ multiplies $y$ by $k^2$ , because $a(kx)^2=k^2\cdot ax^2$ . That is why the area of a square with side $x$ is $S=x^2$ ( $a=1$ ) and the area of a circle with radius $R$ is $S=\pi R^2$ ( $a=\pi$ ): enlarge a shape by a factor of $2$ and its area becomes four times as large.

Functions such as $y=x^2+1$ or $y=(x-1)^2$ do not have the form $y=ax^2$ : their graphs do not pass through $O$ (for $y=x^2+1$ ) or do not have the $y$ -axis as axis of symmetry (for $y=(x-1)^2$ ). In grade 10 you will study general quadratic functions like these.

The graph of $y=ax^2$

Explore: the functions $y=x^2$ and $y=-x^2$

$x$	$-2$	$-1$	$0$	$1$	$2$
$y=x^2$	$4$	$1$	$0$	$1$	$4$
$y=-x^2$	$-4$	$-1$	$0$	$-1$	$-4$

Two opposite values of $x$ give the same value of $y$ , because $(-x)^2=x^2$ . Plotting the points $(x, y)$ on a coordinate grid and joining them with a smooth curve, we get the picture below.

The graphs of

y=x^2

and

y=-x^2

Why is it symmetrical about the $y$ -axis? If the point $(x_0, y_0)$ is on the graph, then $y_0=ax_0^2=a(-x_0)^2$ , so the point $(-x_0, y_0)$ is on the graph too. These two points are reflections of each other in the $y$ -axis. Also $x^2\geq0$ , so $y=ax^2$ has the same sign as $a$ when $x\neq0$ .

Example Drawing a graph

Draw the graph of $y=\dfrac12x^2$ .

Solution. We make a table of values (choosing even $x$ so that $y$ is a whole number):

$x$	$-4$	$-2$	$0$	$2$	$4$
$y=\frac12x^2$	$8$	$2$	$0$	$2$	$8$

We plot the points $(-4, 8)$ , $(-2, 2)$ , $(0, 0)$ , $(2, 2)$ , $(4, 8)$ and join them with a smooth curve.

So the graph is a parabola with vertex $O$ , lying above the $x$ -axis, with the $y$ -axis as its axis of symmetry (see the figure below).

The graph of

y=\frac12x^2

Points on the graph and real-life problems

The point $M(x_0, y_0)$ lies on the graph of $y=ax^2$ exactly when $y_0=ax_0^2$ . So if we know one point of the graph, we can find the coefficient $a$ .

Conversely, if we know the $y$ -coordinate, we find the $x$ -coordinate by solving $ax^2=y_0$ . Because of the symmetry, there are usually two opposite values of $x$ .

Example Answering the opening question

The marble falls from a height of $45$ m, and the distance fallen is $s=5t^2$ (m). How long does it take to reach the ground? After $2$ seconds, how far above the ground is it?

Solution.

Understand. We need the time $t$ when the marble reaches the ground, and the height left when $t=2$ .

Plan. The marble reaches the ground when the distance fallen equals the height, that is $s=45$ . The height left is $45-s$ .

Solve. We have

$5t^2=45\Leftrightarrow t^2=9\Leftrightarrow t=3\ \text{ or }\ t=-3.$

Since $t>0$ , $t=3$ . After $2$ seconds the marble has fallen $s=5\cdot2^2=20$ (m), so it is $45-20=25$ (m) above the ground.

So the marble reaches the ground after $3$ seconds; after $2$ seconds it is $25$ m above the ground.

Look back. Is the result sensible? After $2$ seconds (two thirds of the time) the marble has fallen less than half the height, because it keeps speeding up. If the building were four times as high ( $180$ m), the fall would take only twice as long ( $t^2=36$ , $t=6$ seconds), not four times as long.

Connections. This lesson builds on functions, graphs (grade 8) and square roots. Solving $ax^2=k$ is the simplest case of a quadratic equation in one unknown. A parabola is also the path of a ball thrown at an angle (ignoring air resistance) and the graph of the quadratic functions of grade 10.

Key terms

English	Tiếng Việt
the function $y=ax^2$	hàm số $y=ax^2$
coefficient	hệ số
table of values	bảng giá trị
graph	đồ thị
parabola	parabol
vertex	đỉnh
axis of symmetry	trục đối xứng
lowest / highest point	điểm thấp nhất / cao nhất

Summary

The function $y=ax^2$ ( $a\neq0$ ) is defined for all $x\in\mathbb{R}$ ; $x$ and $-x$ give the same value of $y$ .
Its graph is a parabola with vertex $O$ and axis of symmetry the $y$ -axis.
$a>0$ : the graph is above the $x$ -axis and $O$ is lowest; $a<0$ : the graph is below the $x$ -axis and $O$ is highest.
$M(x_0, y_0)$ is on the graph $\Leftrightarrow y_0=ax_0^2$ ; use this to find $a$ or to find points.
To draw the graph: make a table of values with opposite pairs of $x$ and join with a smooth curve.

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

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

What you need and how to start

What you need: substitute the value of $x$ into the formula, square first, then multiply by $3$ .

Steps:

Work out $x^2$ for each value.
Multiply the result by $3$ .

Show solution

We have $3\cdot(-2)^2=12$ , $3\cdot0^2=0$ , $3\cdot\left(\dfrac13\right)^2=\dfrac13$ and $3\cdot(\sqrt3)^2=9$ .

So the values of $y$ are $12$ , $0$ , $\dfrac13$ and $9$ respectively.

Answer. $12$ ; $0$ ; $\dfrac13$ ; $9$ .

Exercise 2

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

What you need and how to start

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

Steps:

Work out $y$ at each value of $x$ .
Plot the points and join them with a smooth curve.

Show solution

We have the table of values:

$x$	$-4$	$-2$	$0$	$2$	$4$
$y$	$-8$	$-2$	$0$	$-2$	$-8$

So the graph is a parabola with vertex $O$ through these points, lying below the $x$ -axis with the $y$ -axis as its axis of symmetry; $O$ is its highest point.

Answer. A parabola with vertex $O$ through the points in the table, lying below the $x$ -axis.

Exercise 3

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

What you need and how to start

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

Steps:

Work out $2x_0^2$ for each point and compare it with $y_0$ .

Show solution

We have $2\cdot1^2=2$ , $2\cdot(-2)^2=8$ , $2\cdot3^2=18$ and $2\cdot(-1)^2=2\neq-2$ .

So the points $A$ , $B$ , $C$ lie on the graph and the point $D$ does not.

Answer. $A$ , $B$ , $C$ .

Exercise 4

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

a) Find $a$ .

b) Find the points on the graph whose $y$ -coordinate is $24$ .

What you need and how to start

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

Steps:

Solve $9a=6$ .
Solve $ax^2=24$ .

Show solution

a) Substituting $x=-3$ , $y=6$ , we get $9a=6$ , so $a=\dfrac23$ .

b) We have $\dfrac23x^2=24\Leftrightarrow x^2=36\Leftrightarrow x=6$ or $x=-6$ .

So $a=\dfrac23$ , and the required points are $(6, 24)$ and $(-6, 24)$ .

Answer. a) $a=\dfrac23$ ; b) $(6, 24)$ , $(-6, 24)$ .

Exercise 5

An object is dropped from a height of $125$ m. The distance it has fallen after $t$ seconds is $s=5t^2$ (m).

a) After $3$ seconds, how far above the ground is the object?

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

What you need and how to start

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

Steps:

Work out $s$ when $t=3$ .
Solve $5t^2=125$ with $t>0$ .

Show solution

a) When $t=3$ , we have $s=5\cdot3^2=45$ (m). Hence the object is $125-45=80$ (m) above the ground.

b) We have $5t^2=125\Leftrightarrow t^2=25$ . Since $t>0$ , $t=5$ .

So after $3$ seconds the object is $80$ m above the ground, and it reaches the ground after $5$ seconds.

Answer. a) $80$ m; b) $5$ seconds.

Exercise 6

In grade 9 science (the lesson on kinetic and potential energy), the kinetic energy of an object of mass $m$ (kg) moving at speed $v$ (m/s) is $W=\dfrac12mv^2$ (J). Suppose a ball has mass $0.4$ kg; then $W=0.2v^2$ .

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

b) The ball has kinetic energy $45$ J. What is its speed?

What you need and how to start

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

Steps:

Substitute $v=10$ .
Solve $0.2v^2=45$ .

Show solution

a) We have $W=0.2\cdot10^2=20$ (J).

b) We have $0.2v^2=45\Leftrightarrow v^2=225$ . Since $v>0$ , $v=15$ .

So the kinetic energy is $20$ J, and the required speed is $15$ m/s.

Remark. Multiplying the speed by $1.5$ multiplies the kinetic energy by $1.5^2=2.25$ ; a vehicle going twice as fast has four times the kinetic energy.

Answer. a) $20$ J; b) $15$ m/s.

Exercise 7

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

a) Find the $x$ -coordinates of the points on the graph whose $y$ -coordinate is $6$ .

b) Find the value of $y$ when $x=1-\sqrt2$ .

What you need and how to start

What you need: $x^2=2\Leftrightarrow x=\pm\sqrt2$ ; the identity $(a-b)^2=a^2-2ab+b^2$ .

Steps:

Solve $3x^2=6$ .
Expand $(1-\sqrt2)^2$ , then multiply by $3$ .

Show solution

a) We have $3x^2=6\Leftrightarrow x^2=2\Leftrightarrow x=\sqrt2$ or $x=-\sqrt2$ .

b) We have $(1-\sqrt2)^2=1-2\sqrt2+2=3-2\sqrt2$ , so $y=3(3-2\sqrt2)=9-6\sqrt2$ .

So a) $x=\pm\sqrt2$ ; b) $y=9-6\sqrt2$ .

Answer. a) $x=\sqrt2$ , $x=-\sqrt2$ ; b) $9-6\sqrt2$ .

Exercise 8

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

a) Find $a$ .

b) Draw $(P)$ .

c) Find the points on $(P)$ whose $y$ -coordinate is twice their $x$ -coordinate.

What you need and how to start

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

Steps:

Solve $4a=2$ .
Make a table of values with $x\in\{-4, -2, 0, 2, 4\}$ .
The required points have the form $(x, 2x)$ : solve $\dfrac12x^2=2x$ .

Show solution

a) Substituting $x=2$ , $y=2$ , we get $4a=2$ , so $a=\dfrac12$ .

b) $(P)$ is the parabola $y=\dfrac12x^2$ drawn in the example "Drawing a graph".

c) The required points have the form $(x, 2x)$ . Substituting into $(P)$ , we have

$\frac12x^2=2x\Leftrightarrow x^2-4x=0\Leftrightarrow x(x-4)=0\Leftrightarrow x=0\ \text{ or }\ x=4.$

So $a=\dfrac12$ , and the required points are $(0, 0)$ and $(4, 8)$ .

Answer. a) $a=\dfrac12$ ; c) $(0, 0)$ and $(4, 8)$ .

Exercise 9 Challenge

Suppose an entrance arch has the shape of a parabola. It is $8$ m wide at the base and its top is $6$ m above the ground. Choose axes with the origin $O$ at the top of the arch and the $y$ -axis pointing upwards; then the arch is part of the graph of $y=ax^2$ . A lorry $3$ m wide and $5$ m high drives through the middle of the arch. Can the lorry get through?

What you need and how to start

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

Steps:

Find $a$ from the point $(4, -6)$ .
The sides of the lorry are $1.5$ m from the centre line: find $y$ at $x=1.5$ .
The height of the arch there is $6+y$ ; compare it with $5$ .

Show solution

The foot $(4, -6)$ is on the graph, so $16a=-6$ , hence $a=-\dfrac38$ .

At $x=1.5$ , we have $y=-\dfrac38\cdot1.5^2=-\dfrac{27}{32}\approx-0.84$ . Hence the height of the arch above the sides of the lorry is $6-0.84375=5.15625$ (m) $>5$ m.

So the lorry can get through the arch.

Answer. Yes (the arch is about $5.16$ m high above the sides of the lorry).

Mini test

Solutions and answers

Try each exercise on your own first, then check here.

Exercise 1.

We have $3\cdot(-2)^2=12$ , $3\cdot0^2=0$ , $3\cdot\left(\dfrac13\right)^2=\dfrac13$ and $3\cdot(\sqrt3)^2=9$ .

So the values of $y$ are $12$ , $0$ , $\dfrac13$ and $9$ respectively.

Answer. $12$ ; $0$ ; $\dfrac13$ ; $9$ .

Exercise 2.

We have the table of values:

$x$	$-4$	$-2$	$0$	$2$	$4$
$y$	$-8$	$-2$	$0$	$-2$	$-8$

So the graph is a parabola with vertex $O$ through these points, lying below the $x$ -axis with the $y$ -axis as its axis of symmetry; $O$ is its highest point.

Answer. A parabola with vertex $O$ through the points in the table, lying below the $x$ -axis.

Exercise 3.

We have $2\cdot1^2=2$ , $2\cdot(-2)^2=8$ , $2\cdot3^2=18$ and $2\cdot(-1)^2=2\neq-2$ .

So the points $A$ , $B$ , $C$ lie on the graph and the point $D$ does not.

Answer. $A$ , $B$ , $C$ .

Exercise 4.

a) Substituting $x=-3$ , $y=6$ , we get $9a=6$ , so $a=\dfrac23$ .

b) We have $\dfrac23x^2=24\Leftrightarrow x^2=36\Leftrightarrow x=6$ or $x=-6$ .

So $a=\dfrac23$ , and the required points are $(6, 24)$ and $(-6, 24)$ .

Answer. a) $a=\dfrac23$ ; b) $(6, 24)$ , $(-6, 24)$ .

Exercise 5.

a) When $t=3$ , we have $s=5\cdot3^2=45$ (m). Hence the object is $125-45=80$ (m) above the ground.

b) We have $5t^2=125\Leftrightarrow t^2=25$ . Since $t>0$ , $t=5$ .

So after $3$ seconds the object is $80$ m above the ground, and it reaches the ground after $5$ seconds.

Answer. a) $80$ m; b) $5$ seconds.

Exercise 6.

a) We have $W=0.2\cdot10^2=20$ (J).

b) We have $0.2v^2=45\Leftrightarrow v^2=225$ . Since $v>0$ , $v=15$ .

So the kinetic energy is $20$ J, and the required speed is $15$ m/s.

Remark. Multiplying the speed by $1.5$ multiplies the kinetic energy by $1.5^2=2.25$ ; a vehicle going twice as fast has four times the kinetic energy.

Answer. a) $20$ J; b) $15$ m/s.

Exercise 7.

a) We have $3x^2=6\Leftrightarrow x^2=2\Leftrightarrow x=\sqrt2$ or $x=-\sqrt2$ .

b) We have $(1-\sqrt2)^2=1-2\sqrt2+2=3-2\sqrt2$ , so $y=3(3-2\sqrt2)=9-6\sqrt2$ .

So a) $x=\pm\sqrt2$ ; b) $y=9-6\sqrt2$ .

Answer. a) $x=\sqrt2$ , $x=-\sqrt2$ ; b) $9-6\sqrt2$ .

Exercise 8.

a) Substituting $x=2$ , $y=2$ , we get $4a=2$ , so $a=\dfrac12$ .

b) $(P)$ is the parabola $y=\dfrac12x^2$ drawn in the example "Drawing a graph".

c) The required points have the form $(x, 2x)$ . Substituting into $(P)$ , we have

$\frac12x^2=2x\Leftrightarrow x^2-4x=0\Leftrightarrow x(x-4)=0\Leftrightarrow x=0\ \text{ or }\ x=4.$

So $a=\dfrac12$ , and the required points are $(0, 0)$ and $(4, 8)$ .

Answer. a) $a=\dfrac12$ ; c) $(0, 0)$ and $(4, 8)$ .

Exercise 9.

The foot $(4, -6)$ is on the graph, so $16a=-6$ , hence $a=-\dfrac38$ .

At $x=1.5$ , we have $y=-\dfrac38\cdot1.5^2=-\dfrac{27}{32}\approx-0.84$ . Hence the height of the arch above the sides of the lorry is $6-0.84375=5.15625$ (m) $>5$ m.

So the lorry can get through the arch.

Answer. Yes (the arch is about $5.16$ m high above the sides of the lorry).

Mini test: answers

Part I, question 1: B. $a=-\dfrac12\neq0$ . A is a linear function, C has an extra term $2$ , and D has $a=0$ .
Part I, question 2: B. $y=-2\cdot9=-18$ . A has the wrong sign, C squares the $-2$ as well, and D forgets to square.
Part I, question 3: A. $4a=8$ , so $a=2$ . B takes $(-2)^2=-4$ , C takes $8\div2$ , and D takes $8\div(-2)$ .
Part I, question 4: C. At $x=3$ , $y=-9\neq9$ .
Part II, question 1: a) True; b) True; c) False; d) True. a) True: $a>0$ . b) True: $\dfrac13\cdot9=3$ . c) False: at $x=-3$ , $y=3$ . d) True: $x^2=36$ .
Part III, question 1: 4. $5t^2=80\Leftrightarrow t^2=16$ , and $t>0$ , so $t=4$ .
Part III, question 2: 0.25. $16a=4$ , so $a=0.25$ .

The function $y=ax^2$ ( $a\neq0$ )

Explore: distance fallen over time

The graph of $y=ax^2$

Explore: the functions $y=x^2$ and $y=-x^2$

Points on the graph and real-life problems

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

Solutions and answers

Mini test: answers

a) $O(0, 0)$ is the lowest point of $(P)$ .	True	False
b) $(P)$ passes through the point $(3, 3)$ .	True	False
c) $(P)$ passes through the point $(-3, -3)$ .	True	False
d) The points of $(P)$ with $y$ -coordinate $12$ have $x$ -coordinates $6$ and $-6$ .	True	False

The function y = ax² (a ≠ 0)

The function y=ax2y=ax^2y=ax2 (a≠0a\neq0a=0)

Explore: distance fallen over time

The graph of y=ax2y=ax^2y=ax2

Explore: the functions y=x2y=x^2y=x2 and y=−x2y=-x^2y=−x2

Points on the graph and real-life problems

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

The function $y=ax^2$ ( $a\neq0$ )

The graph of $y=ax^2$

Explore: the functions $y=x^2$ and $y=-x^2$