2.13 - Sketching linear and quadratic graphs

Turn a line or quadratic rule into a clear sketch with its important coordinates labelled. You will use intercepts, the sign of a leading coefficient, symmetry and completed squares to show a graph's structure and interpret what it tells you.

Lines, gradients and intercepts

Can two different parabolas cross the $x$ -axis at exactly the same two points? By the end of this lesson you will know why the answer is yes, and what else a sketch needs to distinguish them. Start with the simpler case of a straight line.

You already know

A point $(x,y)$ lies on a graph when its coordinates satisfy the equation. Set $x=0$ to locate a point on the $y$ -axis; set $y=0$ to locate a point on the $x$ -axis. A function's domain contains its inputs and its range contains its outputs.

A sketch shows the correct shape and important features without requiring every point to be plotted to scale. A drawing on a grid uses the scale accurately. Both need labelled axes and the coordinates requested by the question.

Intercept

An intercept is a point where a graph meets a coordinate axis. An $x$ -intercept has $y=0$ ; a $y$ -intercept has $x=0$ . For example, $(3,0)$ is on the $x$ -axis. The point $(0,3)$ is on the $y$ -axis, so interchanging the coordinates changes the intercept.

In a straight-line rule, the coefficient of $x$ tells you how much $y$ changes when $x$ increases by $1$ .

Straight-line features

For a non-vertical straight line,

y=mx+c

the gradient is $m$ , and the $y$ -intercept is $(0,c)$ . If $m\ne0$ , the $x$ -intercept is $\left(-\frac cm,0\right)$ . If $m=0$ , the line is horizontal; it meets the $x$ -axis only when $c=0$ , when the whole line is that axis.
Not on the formulae sheet: learn it.

The sheet supplies a different useful form, $y-y_1=m(x-x_1)$ , for a line through a known point. You will use that in coordinate geometry. Here, use intercepts to anchor a sketch.

Worked example 1: a rising line

Question. Sketch $y=2x-4$ , labelling its intersections with the axes.

Plan. Read the gradient, set each coordinate to zero in turn and join the intercepts with a straight line.

\begin{aligned} m&=2 && \text{positive gradient: line rises} \\ x=0&\quad\Longrightarrow\quad y=-4 && y\text{-intercept} \\ y=0&\quad\Longrightarrow\quad 2x-4=0 && x\text{-intercept} \\ x&=2 && \text{solve the linear equation} \end{aligned}

Why this step? Two different points fix one straight line; the intercepts also provide the labels the question needs.

Answer. Draw a rising straight line through $(0,-4)$ and $(2,0)$ , extending beyond both points, as shown below in blue.

Check. At $x=1$ , the rule gives $y=-2$ , halfway between the two labelled intercepts.

Worked example 2 changes only the sign of the gradient. The line falls instead of rising, while its $y$ -intercept stays fixed.

Worked example 2: a falling line

Question. Sketch $y=-2x-4$ , labelling its intersections with the axes.

Plan. Use the same intercept method, retaining the negative coefficient.

\begin{aligned} m&=-2 && \text{negative gradient: line falls} \\ x=0&\quad\Longrightarrow\quad y=-4 && y\text{-intercept is unchanged} \\ y=0&\quad\Longrightarrow\quad -2x-4=0 && x\text{-intercept} \\ x&=-2 && \text{divide }-2x=4\text{ by }-2 \end{aligned}

Why this step? The negative gradient changes which side of the $y$ -axis contains the zero of the line.

Answer. Draw a falling straight line through $(0,-4)$ and $(-2,0)$ , shown below in clay.

Check. At $x=-1$ , $y=-2$ , halfway between its intercepts.

The picture compares the two lines on the same scales. To draw accurately on a grid, plot the labelled points at their actual coordinates, then use a ruler.

[DIAGRAM: asset_name: Gradient sign and line intercepts; asset_slug: l2fm-line-slope-sign; caption: Changing the gradient sign keeps the y-intercept but moves the x-intercept; description: A blue line y equals 2x minus 4 and a clay line y equals minus 2x minus 4 meet at (0, minus 4). Their x-intercepts are labelled (2, 0) and (minus 2, 0). Axes and a coordinate grid show one line rising and the other falling.]
Changing the gradient sign keeps the y-intercept but moves the x-intercept

A vertical line such as $x=-2$ has undefined gradient and cannot be written as $y=mx+c$ . This is why the straight-line feature rule specifies non-vertical lines.

If both intercepts are the origin, choose one extra point: $y=2x$ passes through $(0,0)$ and $(1,2)$ . A single point would not fix the line's direction.

Watch out

Wrong: the $y$ -intercept of $y=2x-4$ is $(-4,0)$ . Right: it is $(0,-4)$ , because $x=0$ on the $y$ -axis. Write coordinate pairs, not unlabelled numbers.

Now use a new line. The sign of the constant changes which side of the origin holds its vertical intercept.

Quadratic shape and axis intercepts

A quadratic graph is a smooth parabola; its leading coefficient controls whether it opens upwards or downwards, while solving its equation gives its axis crossings.

For $y=ax^2+bx+c$ , with $a\ne0$ , setting $x=0$ gives $(0,c)$ . Setting $y=0$ gives a quadratic equation, whose real solutions locate the $x$ -intercepts. There may be two, one or none. We will distinguish those cases using the turning point in the next part.

Quadratic shape and intercepts

For $y=ax^2+bx+c$ , with $a\ne0$ ,

\begin{aligned} a>0&:\text{ opens upwards} \\ a<0&:\text{ opens downwards} \\ x=0&:\text{ gives the }y\text{-intercept }(0,c) \\ y=0&:\text{ solve }ax^2+bx+c=0 \end{aligned}

Only real roots produce points on the coordinate graph.
Not on the formulae sheet: learn it.

Factorised form is particularly helpful: its zero factors reveal the $x$ -intercepts immediately. It does not make the graph a pair of straight lines; the product still gives a quadratic curve.

Worked example 3: intercepts of an upward parabola

Question. Work out all axis intercepts of $y=x^2-2x-3$ , and state whether it opens upwards or downwards.

Plan. Substitute $x=0$ , then factorise the equation obtained from $y=0$ .

\begin{aligned} x=0&\quad\Longrightarrow\quad y=-3 && \text{vertical intercept} \\ x^2-2x-3&=0 && \text{horizontal intercepts} \\ (x-3)(x+1)&=0 && \text{factorise} \\ x&=3\text{ or }-1 && \text{zero-product rule} \end{aligned}

Why this step? The equation $y=0$ asks where the curve meets the horizontal axis, so both roots belong on that axis.

Answer. The intercepts are $(0,-3)$ , $(3,0)$ and $(-1,0)$ . The graph opens upwards because $a=1>0$ .

Check. Substitution gives $9-6-3=0$ and $1+2-3=0$ at the proposed roots.

Worked example 4 multiplies the whole rule by $-1$ . The two roots stay the same, but the non-zero outputs reverse sign.

Worked example 4: the same roots with a different shape

Question. Work out all axis intercepts of $y=-x^2+2x+3$ , and state whether it opens upwards or downwards.

Plan. Read the new vertical intercept and take out a negative factor before solving $y=0$ .

\begin{aligned} x=0&\quad\Longrightarrow\quad y=3 && \text{vertical intercept} \\ -(x^2-2x-3)&=0 && \text{horizontal intercepts} \\ -(x-3)(x+1)&=0 && \text{factorise} \\ x&=3\text{ or }-1 && \text{the factor }-1\text{ is not zero} \end{aligned}

Why this step? Multiplying a value by $-1$ changes its sign, but zero remains zero, so the $x$ -intercepts do not move.

Answer. The intercepts are $(0,3)$ , $(3,0)$ and $(-1,0)$ . The graph opens downwards because $a=-1<0$ .

Check. At $x=0$ , the two examples give outputs $-3$ and $3$ ; at both roots, each still gives $0$ .

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Geometry

2.13 - Sketching linear and quadratic graphs

Lines, gradients and intercepts

Quadratic shape and axis intercepts

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