Graphical/numerical solutions | Year 10

01 / LEARN

The idea, step by step

Solve by looking for the same height

To solve f(x)=0, look for x-axis intersections. To solve f(x)=g(x), look where the two graphs meet. The solution is an x-coordinate; a full intersection also includes its y-coordinate.

You can instead graph f(x)-g(x) and look for its roots. This often gives a simpler numerical table.

Narrow an interval rather than guess

For a continuous polynomial, opposite signs at two endpoints guarantee at least one root between them. Evaluate smaller intervals until the requested rounding is secure.

A sketch provides an estimate, not an exact surd. Use the algebraic method when an exact answer is required. A sign change is useful evidence, but a repeated root may touch zero without changing sign.

A common trap

Reporting the y-coordinate of an intersection as the answer for x.

Instead: Read the question carefully: solving for x needs horizontal coordinates, while an intersection point needs both coordinates.

02 / WORKED EXAMPLES

See the thinking

Example 1

A positive root to two decimals

Estimate the positive solution of $x^{2} = 2$ to two decimal places.

A sign change brackets the positive root.
x	1.40	1.41	1.42
x²−2	−0.0400	−0.0119	0.0164

Use $f (x) = x^{2} - 2$ . Values at $1.41$ and $1.42$ are $- 0.0119$ and $0.0164$ .
The root is between $1.41$ and $1.42$ . Checking $1.414$ and $1.415$ gives $- 0.000604$ and $0.002225$ .
Thus it rounds to $1.41$ to two decimal places.

x \approx 1.41

Example 2

Intersect a line and a parabola

Find the intersections of $y = x^{2}$ and $y = 2 x + 3$ .

Equal heights require $x^{2} = 2 x + 3$ .
$x^{2} - 2 x - 3 = (x - 3) (x + 1) = 0$ .
For $x = - 1, 3$ , the y-values are $1, 9$ .

(- 1, 1)

and

(3, 9)

Check your understanding

What equation locates intersections of $y = x^{2} - 1$ and $y = x + 1$ ?

Show answer

$x^{2} - 1 = x + 1$ , or $x^{2} - x - 2 = 0$ .

At an intersection, both rules give the same y-value.

03 / WORKSHEET

Your turn

6 questions · 16 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Use the table to locate the positive root of $x^{2} - 3 = 0$ between two listed values.
2 marks
x 1.6 1.7 1.8
x²−3 −0.44 −0.11 0.24
Estimate both roots of $x^{2} - 5 = 0$ to two decimal places.
2 marks
Find the intersections of $y = x^{2} - 2$ and $y = x$ .
3 marks
Use the supplied values to round the positive root of $x^{2} + x - 1 = 0$ to two decimal places.
3 marks
x 0.618 0.619
x²+x−1 −0.000076 0.002161
Solve $x^{2} = 2 x + 1$ exactly, then give decimal answers to two places.
3 marks
Why does a sign-change search miss the root of $f (x) = {(x - 2)}^{2}$ if it checks only x=1 and x=3?
3 marks

x	1.6	1.7	1.8
x²−3	−0.44	−0.11	0.24

x	0.618	0.619
x²+x−1	−0.000076	0.002161

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Use the table to locate the positive root of $x^{2} - 3 = 0$ between two listed values.
2 marks
x 1.6 1.7 1.8
x²−3 −0.44 −0.11 0.24
$f (1.7) = - 0.11$ and $f (1.8) = 0.24$ have opposite signs.
$1.7 < x < 1.8$
Estimate both roots of $x^{2} - 5 = 0$ to two decimal places.
2 marks
The exact roots are $\pm \sqrt{5} \approx \pm 2.236068$ .
$x \approx - 2.24, 2.24$
Find the intersections of $y = x^{2} - 2$ and $y = x$ .
3 marks
$x^{2} - x - 2 = (x - 2) (x + 1) = 0$ ; then $y = x$ .
$(- 1, - 1)$ and $(2, 2)$
Use the supplied values to round the positive root of $x^{2} + x - 1 = 0$ to two decimal places.
3 marks
x 0.618 0.619
x²+x−1 −0.000076 0.002161
The root is between $0.618$ and $0.619$ , so it rounds to $0.62$ .
$x \approx 0.62$
Solve $x^{2} = 2 x + 1$ exactly, then give decimal answers to two places.
3 marks
Complete the square: ${(x - 1)}^{2} = 2$ .
$x = 1 \pm \sqrt{2} \approx - 0.41, 2.41$
Why does a sign-change search miss the root of $f (x) = {(x - 2)}^{2}$ if it checks only x=1 and x=3?
3 marks
$f (1) = f (3) = 1$ , but the curve touches the axis at its repeated root $2$ .
Both endpoint values are positive even though f(2)=0.

x	1.6	1.7	1.8
x²−3	−0.44	−0.11	0.24

x	0.618	0.619
x²+x−1	−0.000076	0.002161

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Curriculum reference & lesson scope

Australian Curriculum v9.0 and South Australian Curriculum

Year 10 quadratics, including the fuller school-course algebra requested for Ascent: non-monic factorisation, completing the square and the quadratic formula. These lessons consolidate Year 9 monic quadratics and extend the available solving methods.

Official curriculum source ↗

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