Graphical/numerical solutions | Year 10

01 / LEARN

Understand the idea.

Solve by looking for the same height

A graphical solution is an input where a target relationship holds. For $f (x) = 0$ , read horizontal intercepts; for $f (x) = g (x)$ , read intersections, where the two rules give equal outputs.

Subtracting the rules produces the equivalent equation $f (x) - g (x) = 0$ . Its roots have the same input coordinates as the intersections, although its graph does not share their original heights.

Narrow an interval rather than guess

Polynomials are continuous, so opposite signs at interval endpoints guarantee at least one root between them. Evaluate closer inputs to narrow that interval until every possible value rounds to the same requested result.

A sign change does not show every root: a repeated root can touch zero and turn back without changing sign. Graphical estimates and numerical brackets are approximations; keep an exact algebraic expression when an exact answer is required.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

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

Worked solution

1
Create a zero-finding function.
$f (x) = x^{2} - 2, x > 0$
The required positive solution is the positive root of this continuous polynomial.
2
Bracket the positive root.
$\begin{aligned} f (1.41) & = 1.9881 - 2 = - 0.0119 \\ f (1.42) & = 2.0164 - 2 = 0.0164 \end{aligned}$
The sign change places a root between $1.41$ and $1.42$ ; squaring increases for positive inputs.
3
Refine around the rounding boundary.
$\begin{aligned} f (1.414) & = 1.999396 - 2 = - 0.000604 \\ f (1.415) & = 2.002225 - 2 = 0.002225 \end{aligned}$
The root lies above $1.414$ and strictly below $1.415$ , the boundary for rounding up to $1.42$ .
4
Round only after securing the interval.
$1.414 < x < 1.415 ⟹ x \approx 1.41$
Every value in this interval rounds to the same hundredth, so the two-decimal estimate is justified.

Answer

$x \approx 1.41$

EXAMPLE 02

Intersect a line and a parabola

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

Worked solution

1
Set the heights equal.
$x^{2} = 2 x + 3$
An intersection point satisfies both rules with the same input and output.
2
Rearrange and factor.
$\begin{aligned} x^{2} - 2 x - 3 & = 0 \\ (x - 3) (x + 1) & = 0 \end{aligned}$
The constants $- 3$ and $1$ multiply to $- 3$ and add to $- 2$ .
3
Solve for both horizontal coordinates.
$x = 3 or x = - 1$
Each zero factor gives one intersection input.
4
Calculate and check the common heights.
$\begin{aligned} x = - 1 : & y = {(- 1)}^{2} = 1 = 2 (- 1) + 3 \\ x = 3 : & y = 3^{2} = 9 = 2 (3) + 3 \end{aligned}$
Both rules agree, so the intersections are $(- 1, 1)$ and $(3, 9)$ .

Answer

$(- 1, 1)$ and $(3, 9)$

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

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

Equal heights require $x^{2} - 1 = x + 1$ . Subtracting $x + 1$ from both sides gives the equivalent equation $x^{2} - x - 2 = 0$ , whose roots locate the intersection inputs.

Keep practising

6 worksheet questions · 16 marks

Open worksheet 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
$x^{2} - 3$ −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
$x^{2} + x - 1$ −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
$x^{2} - 3$	−0.44	−0.11	0.24

$x$	0.618	0.619
$x^{2} + x - 1$	−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
$x^{2} - 3$ −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
$x^{2} + x - 1$ −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
$x^{2} - 3$	−0.44	−0.11	0.24

$x$	0.618	0.619
$x^{2} + x - 1$	−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 ↗

In this lesson

Understand the idea.

Solve by looking for the same height

Narrow an interval rather than guess

See the method in action.

A positive root to two decimals

Create a zero-finding function.

Bracket the positive root.

Refine around the rounding boundary.

Round only after securing the interval.

Intersect a line and a parabola

Set the heights equal.

Rearrange and factor.

Solve for both horizontal coordinates.

Calculate and check the common heights.

Try it yourself.

Check your understanding

Keep practising