Linear and Quadratic Functions

Lessons

Senior 3 · Mathematics

Linear and Quadratic Functions

5 lessons

Lesson 5 of 5

Sketching parabolas and solving equations graphically

25 min

A sketch, not a plot

A table of values gives an accurate graph, but it takes time. In an exam you are often asked for a sketch: the right shape, with the key points marked and labelled exactly. Five facts are enough to fix a parabola. The same graph can then solve equations: wherever the curve meets a line, the coordinates satisfy both equations at once.

Important

Sketching $y = a x^{2} + b x + c$ in five steps

Shape: $\cup$ if $a > 0$ , $\cap$ if $a < 0$ .
$y$ -intercept: put $x = 0$ : the point $(0, c)$ .
$x$ -intercepts: solve $a x^{2} + b x + c = 0$ (Unit 2). There may be two, one or none.
Vertex and axis: $x = -\frac{b}{2a}$ , then substitute.
Mirror point: reflect $(0, c)$ in the axis to get a second point at the same height.

Worked example 1: a full sketch

Sketch $y = x^{2} - 6 x + 8$ .

Solution $a = 1 > 0$ : $\cup$ . $y$ -intercept $(0, 8)$ .

$x$ -intercepts: $x^{2} - 6 x + 8 = (x - 2) (x - 4) = 0$ , so $(2, 0)$ and $(4, 0)$ .

Axis $x = -\frac{-6}{2} = 3$ (midway between 2 and 4 ✓); vertex $y = 9 - 18 + 8 = - 1$ , so $(3, - 1)$ .

Mirror of $(0, 8)$ in $x = 3$ : $(6, 8)$ .

Five labelled points and the axis fix the sketch.

Important

How many $x$ -intercepts? Ask the discriminant. The discriminant of $a x^{2} + b x + c$ is $\Delta = b^2 - 4ac$ (Unit 2).

$\Delta$	Roots of $a x^{2} + b x + c = 0$	The parabola and the $x$ -axis
$\Delta > 0$	two different real roots	crosses it twice
$\Delta = 0$	one repeated root	touches it at the vertex
$\Delta < 0$	no real roots	never meets it

Same axis, different c: the discriminants are 4, 0 and −8.

Definition

Solving an equation graphically: the solutions of $f (x) = k$ are the $x$ -coordinates of the points where the curve $y = f (x)$ meets the horizontal line $y = k$ . In particular, the roots of $f (x) = 0$ are the $x$ -intercepts.

Example: the roots of $x^{2} - 6 x + 8 = 0$ are read from the graph above as $x = 2$ and $x = 4$ .

Worked example 2: one graph, three equations

Use the graph of $y = x^{2} - 2 x - 3$ to solve (a) $x^{2} - 2 x - 3 = 0$ (b) $x^{2} - 2 x - 3 = 5$ (c) $x^{2} - 2 x = 1$ .

Solution (a) Read the $x$ -intercepts: $x = - 1$ or $x = 3$ .

(b) Draw $y = 5$ . It meets the curve at $x = - 2$ and $x = 4$ . Check: $16 - 8 - 3 = 5$ ✓.

(c) First make the left side match the graph: subtract 3 from both sides, $x^{2} - 2 x - 3 = - 2$ . Draw $y = - 2$ : it meets the curve at $x \approx -0.4$ and $x \approx 2.4$ . (The exact roots are $1 \pm \sqrt{2} \approx -0.41$ and $2.41$ .)

The dashed lines drop from each meeting point to the x-axis, where the solutions are read.

Important

Where a line meets a parabola. To find where $y = m x + k$ meets $y = a x^{2} + b x + c$ , set the two expressions for $y$ equal:

ax^2 + bx + c = mx + k \quad\Longrightarrow\quad ax^2 + (b - m)x + (c - k) = 0.

Each root gives the $x$ -coordinate of a meeting point; substitute into the line to get $y$ . The discriminant of this new equation tells you how many points there are: two if $\Delta > 0$ , one if $\Delta = 0$ (the line is a tangent, touching the curve), and none if $\Delta < 0$ .

Worked example 3: two meeting points

Find where the line $y = x + 1$ meets the parabola $y = x^{2} - 1$ .

Solution

x^2 - 1 = x + 1 \;\Longrightarrow\; x^2 - x - 2 = 0 \;\Longrightarrow\; (x - 2)(x + 1) = 0.

$x = 2$ gives $y = 3$ ; $x = - 1$ gives $y = 0$ . The points are $(2, 3)$ and $(- 1, 0)$ . Check $(2, 3)$ in the parabola: $4 - 1 = 3$ ✓.

The line crosses the parabola twice.

Worked example 4: a line that just touches

Find the value of $c$ for which the line $y = 2 x + c$ touches the parabola $y = x^{2}$ , and the point of contact.

Solution $x^{2} = 2 x + c$ gives $x^{2} - 2 x - c = 0$ . The line touches when this has a repeated root, that is, $\Delta = 0$ :

(-2)^2 - 4(1)(-c) = 4 + 4c = 0 \quad\Longrightarrow\quad c = -1.

Then $x^{2} - 2 x + 1 = (x - 1)^{2} = 0$ , so $x = 1$ and $y = 1$ : the tangent $y = 2 x - 1$ touches at $(1, 1)$ . For $c > - 1$ the line cuts the curve twice; for $c < - 1$ it misses.

Warning

Match the graph before you read it. To solve $x^{2} - 2 x = 1$ with the graph of $y = x^{2} - 2 x - 3$ , you must draw $y = - 2$ , not $y = 1$ : both sides must first be changed so that the left side is exactly the function drawn.

Readings are estimates. A graph gives roots to about one decimal place. When the question says "exact", use algebra.

Tip

Label every key point with its coordinates on a sketch, even when the drawing is rough. The marks are for the intercepts, the vertex and the shape, not for neat curves.

Khan Academy: Quadratic systems, a line and a parabola. Watch how setting the two expressions for y equal gives a quadratic equation, and how its solutions are the meeting points on the graph.

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Practice

The last three are harder.

Sketch $y = x^{2} - 2 x - 8$ , marking the intercepts and the vertex.
Sketch $y = - x^{2} + 2 x + 3$ , marking the intercepts and the vertex.
Use the discriminant to find how many times each parabola meets the $x$ -axis: (a) $y = x^{2} + 2 x + 5$ (b) $y = 4 x^{2} - 12 x + 9$ (c) $y = 2 x^{2} - 3 x - 1$ .
Draw $y = x^{2} - x - 6$ for $-3 \le x \le 4$ . Use it to solve (a) $x^{2} - x - 6 = 0$ (b) $x^{2} - x - 2 = 0$ .
Find where $y = x - 3$ meets $y = x^{2} - 3 x$ .
Show that the line $y = 2 x - 5$ never meets the parabola $y = x^{2}$ .
Find $k$ so that the line $y = x + k$ touches $y = x^{2} + 3 x$ , and give the point of contact.

Show answersHide answers

$\cup$ ; $(0, - 8)$ ; $(x - 4) (x + 2) = 0$ gives $(- 2, 0)$ , $(4, 0)$ ; vertex $(1, - 9)$ .
$\cap$ ; $(0, 3)$ ; $- (x - 3) (x + 1) = 0$ gives $(- 1, 0)$ , $(3, 0)$ ; vertex $(1, 4)$ .
(a) $\Delta = 4 - 20 = -16$ : none (b) $\Delta = 144 - 144 = 0$ : one (c) $\Delta = 9 + 8 = 17$ : two.
(a) $x = - 2$ or $x = 3$ . (b) $x^{2} - x - 2 = 0$ means $x^{2} - x - 6 = - 4$ : draw $y = - 4$ , giving $x = - 1$ or $x = 2$ .
$x^{2} - 3 x = x - 3$ gives $x^{2} - 4 x + 3 = 0$ , so $x = 1$ or $3$ : the points $(1, - 2)$ and $(3, 0)$ .
$x^{2} = 2 x - 5$ gives $x^{2} - 2 x + 5 = 0$ , with $\Delta = 4 - 20 = -16 < 0$ : no real roots, so no meeting point.
$x^{2} + 3 x = x + k$ gives $x^{2} + 2 x - k = 0$ ; $\Delta = 4 + 4k = 0$ , so $k = - 1$ . Then $(x + 1)^{2} = 0$ : contact at $(- 1, - 2)$ .

1A sketch, not a plot

A sketch, not a plot