Sec 2 Linear and Quadratic Graphs: Concepts & Worked Examples

Secondary 2 G3 Mathematics · Study guide

Connect gradient, intercepts, quadratic tables and turning points to graphical solutions and real situations, with four worked examples and an interactive graph.

Concept → worked examples → practice18 min read · practice linked below

Read a graph as a relationship

A graph shows all the pairs $(x,y)$ that satisfy a rule. A straight line has a constant rate of change. A quadratic curve changes direction at a turning point. Learning to move between an equation, a table and a graph helps you solve problems and check answers.

Before you begin: practise plotting ordered pairs, substituting negative numbers with brackets and reading the scale on each axis. The two axes may use different scales.

Gradient and intercepts

For a non-vertical line, $y=mx+c$ . The gradient $m$ is the change in $y$ for each unit increase in $x$ , and $c$ is the y-intercept. Using two points,

$m=\frac{y_2-y_1}{x_2-x_1}.$

Subtract in the same order above and below the fraction. A positive gradient rises from left to right; a negative gradient falls. A horizontal line has gradient zero. A vertical line has equation $x=a$ ; its gradient is undefined because the horizontal change is zero.

Worked example 1 — find a falling line

A line passes through $A(-2,7)$ and $B(4,-2)$ . Find its equation and both intercepts.

The horizontal change from A to B is $4-(-2)=6$ , while the vertical change is $-2-7=-9$ . Thus $m=-9/6=-3/2$ .
Substitute A into $y=-\frac32x+c$ : $7=-\frac32(-2)+c=3+c$ , so $c=4$ .
The equation is $y=-\frac32x+4$ . The y-intercept is the point $(0,4)$ .
At the x-intercept, $y=0$ . Hence $\frac32x=4$ , so the point is $(\frac83,0)$ .

Check: at $x=4$ , the equation gives $-6+4=-2$ , matching B. The negative gradient agrees with the falling line.

Read gradient from coordinate changes, rather than the line’s apparent angle on a resized screen.

Plot a quadratic with a smooth curve

A quadratic has the form $y=ax^2+bx+c$ , with $a\ne0$ . If $a>0$ , it opens upwards and has a minimum. If $a<0$ , it opens downwards and has a maximum. Its vertical line of symmetry passes through the turning point.

Worked example 2 — build the table and locate the minimum

Plot $y=x^2-2x-3$ for $-2\le x\le4$ , and find its minimum and roots.

At $x=-2$ , use brackets: $(-2)^2-2(-2)-3=4+4-3=5$ . Repeating for each x-value gives:

Values for $y=x^2-2x-3$
x	−2	−1	0	1	2	3	4
y	5	0	−3	−4	−3	0	5

Plot every pair and join them with a smooth curve. The equal heights at $x=0$ and $x=2$ lie equally far from $x=1$ . The turning point is $(1,-4)$ , the minimum value is $-4$ , and the line of symmetry is $x=1$ .

The roots solve $x^2-2x-3=0$ , so read where the curve meets the x-axis: $x=-1$ and $x=3$ . Check by substitution: $1+2-3=0$ and $9-6-3=0$ .

The dotted line is the symmetry line x = 1. It is vertical, so its equation begins with x.

Use intersections to solve equations

To find $x$ when $y=k$ , draw the horizontal line $y=k$ . To equate two expressions, graph both on the same axes. Their intersection points have the same output for the same input. Give the x-values if the question asks for solutions; give both coordinates if it asks for intersection points.

Worked example 3 — choose the right straight line

Use $y=x^2-2x-3$ to solve $x^2-3x-4=0$ .

Rearrange to match the plotted expression: $x^2-2x-3=x+1$ . The extra line must be $y=x+1$ .
Plot $(0,1)$ and $(3,4)$ and draw the straight line through them.
Read the two intersections: $(-1,0)$ and $(4,5)$ . Therefore the equation’s solutions are $x=-1$ and $x=4$ .

Independent check: $x^2-3x-4=(x-4)(x+1)$ , so both readings are exact here. For an intersection between grid marks, report an estimate to the precision supported by the graph.

Solid: y = x² − 2x − 3. Dashed: y = x + 1. The shared points solve the rearranged equation.

Interpret the part of the graph that makes sense

Worked example 4 — height, maximum and duration

A ball has height $h=12+8t-2t^2$ metres, $t$ seconds after release. Find its greatest height, estimate when it lands, and find how long it is above 18 m.

The graph opens downwards. Equal heights at $t=0$ and $t=4$ show symmetry about $t=2$ . At that time, $h=12+16-8=20$ . The greatest height is 20 m.
Ground level is $h=0$ . The positive intercept is $t\approx5.16$ s, so the ball lands at about 5.2 s to 1 decimal place. The negative intercept is before release and is excluded.
Draw $h=18$ . It meets the curve at $t=1$ and $t=3$ . The ball is above 18 m for $1<t<3$ , a duration of $3-1=\mathbf{2\text{ s}}$ .

Check: $h(1)=18=h(3)$ , and the peak of 20 m lies above 18 m. The physical model runs from release until the first ground contact; continuing the curve below ground would not describe this flight.

The dashed horizontal line is h = 18 m. Subtract the two crossing times to find a duration.

Explore two, one or no intersections

Move the line

The curve stays at $y=x^2-2x-3$ . Choose the line’s gradient and intercept. Its intersections solve $x^2-2x-3=mx+c$ .

Solid curve and dashed line. Red points mark the intersections; each point supplies one solution for x.

Line y = 0. Two intersections: (−1, 0) and (3, 0). The equation has two solutions for x.

Common mistakes and reliable checks

Using screen steepness as gradient: use coordinate differences and the printed axis scales.
Squaring a negative incorrectly: $(-3)^2=9$ , whereas $-3^2=-9$ . Put substituted values in brackets.
Joining a parabola with straight segments: use a smooth curve, supported by the table and its symmetry.
Giving the wrong type of answer: the minimum value is a y-value; a turning point is an ordered pair; a symmetry line is an equation.
Reading just one crossing: scan the whole stated domain. A horizontal line may cut twice, touch once, or miss the curve.
False precision: label axes and units, retain calculator accuracy during checks, and round only the final result as requested.

Practise with our topic questions

Use the Sec 2 linear and quadratic graphs question bank to apply these methods. Attempt each question before opening its worked solution. Compare your reasoning as well as the final answer.