Reading Values from a Quadratic Graph

How to read $y$ for a given $x$ and $x$ for a given $y$ from the graph of a quadratic function, on any scale and in a setting, and how to check each value by substituting.

Before you start

In chapter 2 you read values from a straight-line graph with dashed lines: up from the $x$ -axis to the line, then across to the $y$ -axis.
In lesson 5.5 you found where a quadratic graph meets the $x$ -axis, and in lesson 5.6 its turning point: a ∪-shaped curve is lowest there, a ∩-shaped curve highest.

Reading y for a Given x

Reading a value of y

Start at the given value on the $x$ -axis. Draw a dashed line straight up (or down) to the curve, then straight across to the $y$ -axis, and read the value there.

Between grid lines, judge the fraction of a square: halfway, a quarter of the way, and so on. A value read between grid lines is an estimate, so write it with $\approx$ . Then check it by substituting: when the check gives exactly that value, the answer is written with $=$ .

Values read from a scale are given as decimals: $1\frac{1}{4}$ is written $1.25$ .

Example 1: Between the grid lines

The graph of $y = x^2 + 2x - 4$ is drawn below. Use it to find the value of $y$ when
(a) $x = -3.5$
(b) $x = 0.5$ .

Solution

(a) From $x = -3.5$ , halfway between $-4$ and $-3$ , go up to the curve and across. The curve is about a quarter of a square above $1$ : $y \approx 1.25$ .

(b) From $x = 0.5$ , go down to the curve and across. It is about three quarters of a square below $-2$ : $y \approx -2.75$ .

Check by substituting: $(-3.5)^2 + 2(-3.5) - 4 = 12.25 - 7 - 4 = 1.25$ ✓ and $0.5^2 + 2(0.5) - 4 = 0.25 + 1 - 4 = -2.75$ ✓. Both readings are exact.

(a) $y = 1.25$
(b) $y = -2.75$

Reading x for a Given y

Going the other way, start on the $y$ -axis and draw a dashed line across. A ∪-shaped curve is met twice by a line above its minimum point, once by the line through its minimum point, and not at all by a line below it. A ∩-shaped curve is the other way up: twice below its maximum point, not at all above it. From each point where the line meets the curve, go down (or up) to the $x$ -axis.

Example 2: Two, one or no values

Use the graph of $y = x^2 + 2x - 4$ in Example 1 to find the values of $x$ for which
(a) $y = 4$
(b) $y = -5$
(c) $y = -6$ .

Solution

(a) The line $y = 4$ meets the curve twice, above $x = -4$ and $x = 2$ . Check: $16 - 8 - 4 = 4$ ✓ and $4 + 4 - 4 = 4$ ✓.

(b) The line $y = -5$ just touches the curve at its minimum point, below $x = -1$ on the $x$ -axis. Check: $1 - 2 - 4 = -5$ ✓.

(a) $x = -4$ or $x = 2$
(b) $x = -1$
(c) there is no such value of $x$ .

Example 3: A scale of 2 units to a square

The graph of $y = 7 + 2x - x^2$ for $-3 \le x \le 5$ is drawn below, with each square on the $y$ -axis standing for 2 units. Find
(a) the value of $y$ when $x = 0$
(b) the values of $x$ when $y = 4$ .

Solution

Read the numbers on the axis, not the squares: one square on the $y$ -axis is $2$ units, so half a square is $1$ unit.

(a) At $x = 0$ the curve is half a square above $6$ : $y \approx 7$ . Check: $7 + 0 - 0 = 7$ ✓, so $y = 7$ exactly.

(b) The value $4$ is on a grid line. The line $y = 4$ meets the curve above $x = -1$ and $x = 3$ . Check: $7 - 2 - 1 = 4$ ✓ and $7 + 6 - 9 = 4$ ✓.

(a) $y = 7$
(b) $x = -1$ or $x = 3$

Reading a Graph from a Setting

In a setting the axes have names and units, and only the part of the curve that makes sense is drawn. Read the values as before, and give each answer with its unit.

Example 4: The height of a ball

A ball is thrown upwards from the ground. Its height, $h$ metres, after $t$ seconds is $h = 25t - 5t^2$ . The graph is drawn below. Find
(a) the times at which the ball is 20 m above the ground
(b) the time at which it lands.

Solution

(a) Draw the line $h = 20$ across. It meets the curve twice, above $t = 1$ (going up) and $t = 4$ (coming down). Check: $25 - 5 = 20$ ✓ and $100 - 80 = 20$ ✓.

(b) The ball lands when $h = 0$ again: the curve meets the $t$ -axis at $t = 5$ . Check, as in lesson 5.5: $25t - 5t^2 = 5t(5 - t) = 0$ gives $t = 0$ (when it is thrown) or $t = 5$ ✓.

(a) after 1 second and after 4 seconds
(b) after 5 seconds

Example 5: Only one value on the part drawn

A stone is dropped from a window 45 m above the ground. Its height, $h$ metres, after $t$ seconds is $h = 45 - 5t^2$ , for $0 \le t \le 3$ . The graph is drawn below. Find
(a) the height of the stone after 1.5 seconds
(b) the time at which it is 25 m above the ground.

Solution

(a) From $t = 1.5$ , go up to the curve and across. The curve is about three quarters of a square above $30$ ; one square is $5$ m, so $h \approx 30 + 3.75 = 33.75\text{.}$ Check: $45 - 5(1.5)^2 = 45 - 5 \times 2.25 = 45 - 11.25 = 33.75$ ✓.

(b) The line $h = 25$ meets the drawn curve once, above $t = 2$ . Check: $45 - 5 \times 4 = 25$ ✓.

The equation $45 - 5t^2 = 25$ is also true for $t = -2$ , but that time is before the stone was dropped, and that part of the curve is not drawn.

(a) $33.75$ m (about 34 m)
(b) after 2 seconds

Common Mistakes

Reading only one value of $x$ for a given $y$ . A line across a ∪-shaped or ∩-shaped curve usually meets it twice.
Counting squares on an axis whose scale is not 1 unit to a square. Read the numbers on the axis.
Writing a value read between grid lines as exact without checking it. Write it with $\approx$ and check by substituting.
Reading off the part of the curve that is not drawn, or giving a value that makes no sense in the setting, such as a negative time.

Practice Questions

Draw your reading lines on the graph, and check each value by substituting. The questions get harder as you go.

Question 1

The graph of $y = x^2 - 4x + 1$ for $-1 \le x \le 5$ is drawn below. Use it to find
(a) $y$ when $x = 3$
(b) $x$ when $y = 6$
(c) $x$ when $y = 1$
(d) $x$ when $y = -4$ .

ShowHide solution

(a) At $x = 3$ : $y = -2$ . Check: $9 - 12 + 1 = -2$ ✓.

(b) The line $y = 6$ meets the curve at $x = -1$ and $x = 5$ . Check: $1 + 4 + 1 = 6$ ✓; $25 - 20 + 1 = 6$ ✓.

(d) The lowest point of the curve is $(2, -3)$ , so the line $y = -4$ is below it and never meets the curve.

(a) $y = -2$
(b) $x = -1$ or $5$
(c) $x = 0$ or $4$
(d) no value of $x$

Question 2

The graph of $y = 4 + 3x - x^2$ for $-2 \le x \le 5$ is drawn below. Use it to find
(a) $y$ when $x = 3.5$
(b) $x$ when $y = 4$
(c) $x$ when $y = 6$
(d) $x$ when $y = 7$ .

ShowHide solution

(a) From $x = 3.5$ the curve is about a quarter of a square above $2$ : $y \approx 2.25$ . Check: $4 + 10.5 - 12.25 = 2.25$ ✓, so $y = 2.25$ .

(b) The line $y = 4$ meets the curve at $x = 0$ and $x = 3$ . Check: $4 + 0 - 0 = 4$ ✓; $4 + 9 - 9 = 4$ ✓.

(d) The highest point is halfway between $x = 1$ and $x = 2$ , which give equal values: at $x = 1.5$ , $y = 4 + 4.5 - 2.25 = 6.25\text{.}$ The line $y = 7$ is above it and never meets the curve.

(a) $y = 2.25$
(b) $x = 0$ or $3$
(c) $x = 1$ or $2$
(d) no value of $x$

Question 3

The graph of $y = x^2 - 6x + 5$ for $0 \le x \le 6$ is drawn below with 2 units to a square on the $y$ -axis. Use it to find
(a) $y$ when $x = 0$
(b) $x$ when $y = -3$
(c) the roots of $x^2 - 6x + 5 = 0$ .

ShowHide solution

(a) At $x = 0$ the curve is half a square above $4$ : $y \approx 5$ . Check: $c = 5$ ✓, so $y = 5$ .

(b) $-3$ is halfway between $-2$ and $-4$ . The line $y = -3$ meets the curve at $x = 2$ and $x = 4$ . Check: $4 - 12 + 5 = -3$ ✓; $16 - 24 + 5 = -3$ ✓.

(a) $y = 5$
(b) $x = 2$ or $x = 4$
(c) $x = 1$ or $x = 5$

Question 4

The graph of $y = 2x^2 - 3x - 5$ for $-1.5 \le x \le 3$ is drawn below with 2 squares to 1 unit on the $x$ -axis. Use it to find
(a) $y$ when $x = 1.5$
(b) $x$ when $y = -3$
(c) $x$ when $y = 4$ .

ShowHide solution

One square on the $x$ -axis is $0.5$ , and the axis is numbered at every whole number.

(a) $x = 1.5$ is one square to the right of $1$ . Down to the curve and across: $y = -5$ . Check: $4.5 - 4.5 - 5 = -5$ ✓.

(b) The line $y = -3$ meets the curve at $x = -0.5$ and $x = 2$ . Check: $0.5 + 1.5 - 5 = -3$ ✓; $8 - 6 - 5 = -3$ ✓.

(a) $y = -5$
(b) $x = -0.5$ or $x = 2$
(c) $x = -1.5$ or $x = 3$

Question 5

Wei Ling uses the graph of $y = x^2 - 2x$ below to find $x$ when $y = 3$ . She writes “ $x = 3$ ”. What has she missed? Give the full answer.

ShowHide solution

The line $y = 3$ is above the lowest point of the ∪-shaped curve, so it meets the curve twice. She has read only the point on the right.

Check: $(-1)^2 - 2(-1) = 1 + 2 = 3$ ✓ and $9 - 6 = 3$ ✓.

She has missed the second value: $x = -1$ or $x = 3$ .

Question 6

A stone is thrown upwards from the top of a cliff 40 m high. Its height above the sea, $h$ metres, after $t$ seconds is $h = 40 + 10t - 5t^2$ . The graph is drawn below. Find
(a) the times when the stone is 40 m above the sea
(b) the time when it reaches the sea.

ShowHide solution

(a) The line $h = 40$ meets the curve at $t = 0$ (when it is thrown) and at $t = 2$ (on the way down). Check: $40 + 20 - 20 = 40$ ✓.

(b) The curve meets the $t$ -axis at $t = 4$ . Check: $40 + 40 - 80 = 0$ ✓.

(a) at $t = 0$ and after 2 seconds
(b) after 4 seconds

Question 7

A rectangle has sides $x$ cm and $(8 - x)$ cm, so its area is $A = 8x - x^2$ square centimetres. The graph is drawn below. Find
(a) the area when $x = 3$
(b) the values of $x$ that give an area of 12 cm².

ShowHide solution

(a) At $x = 3$ the curve is half a square above $14$ : $A \approx 15$ . Check: $24 - 9 = 15$ ✓, so $A = 15$ .

(b) The line $A = 12$ meets the curve at $x = 2$ and $x = 6$ . Check: $16 - 4 = 12$ ✓; $48 - 36 = 12$ ✓. The rectangle is 2 cm by 6 cm either way.

(a) 15 cm²
(b) $x = 2$ or $x = 6$

Question 8

A farmer has 20 m of fencing for three sides of a rectangular pen; a wall makes the fourth side. When each of the two sides at right angles to the wall is $x$ m long, the area of the pen is $A = 20x - 2x^2$ square metres. The graph for $0 \le x \le 10$ is drawn below. Find
(a) the area when $x = 3$
(b) the values of $x$ that give an area of 32 m²
(c) the value of $x$ that gives an area of 50 m².

ShowHide solution

One square on the $A$ -axis is $4$ m², so half a square is $2$ m².

(a) At $x = 3$ the curve is half a square above $40$ : $A \approx 42$ . Check: $60 - 18 = 42$ ✓, so $A = 42$ .

(b) The line $A = 32$ meets the curve at $x = 2$ and $x = 8$ . Check: $40 - 8 = 32$ ✓; $160 - 128 = 32$ ✓.

(c) The line $A = 50$ just touches the curve at its highest point, above $x = 5$ . Check: $100 - 50 = 50$ ✓. The pen is then 5 m by 10 m.

(a) 42 m²
(b) $x = 2$ or $x = 8$
(c) $x = 5$

Key Points

To read $y$ for a given $x$ : up (or down) to the curve, then across to the $y$ -axis.
To read $x$ for a given $y$ : across to the curve, then down (or up) to the $x$ -axis. There may be two values, one (at the turning point) or none.
Read the numbers on each axis; a square need not be 1 unit on either axis.
A value read between grid lines is an estimate: write it with $\approx$ , and check it by substituting.
Only the part of the curve that is drawn can be read; in a setting, keep only the values that make sense.