S3 Mathematics Practice Book · Unit 6: Linear and quadratic functions

Unit 6 Linear and quadratic functions

Key unit competence: Apply graphical behaviour and key features of linear and quadratic functions in solving real-world problems.

Key ideas

Line with gradient $m$ through $(x_{1}, y_{1})$ : $y - y_{1} = m (x - x_{1})$ . Through two points: first $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ .
Parallel lines have equal gradients: $m_{1} = m_{2}$ . Perpendicular lines have $m_{1} \times m_{2} = - 1$ .
A quadratic function $f (x) = a x^{2} + b x + c$ ( $a \neq = 0$ ) has a parabola as its graph. If $a > 0$ it opens upwards (minimum); if $a < 0$ it opens downwards (maximum).
Vertex: $x = - \frac{b}{2 a}$ , then $y = f (- \frac{b}{2 a})$ . The axis of symmetry is the line $x = - \frac{b}{2 a}$ .
Intercepts: y-intercept $(0, c)$ ; x-intercepts are the roots of $a x^{2} + b x + c = 0$ (if any).
To sketch: find the direction of opening, intercepts, vertex and axis, then use a table of values.
Graphical solution of $a x^{2} + b x + c = 0$ : read the x-coordinates where the parabola crosses the x-axis; for $a x^{2} + b x + c = k$ use the line $y = k$ .

Formulas to know

y - y_{1} = m (x - x_{1})

m_{1} = m_{2} (∥), m_{1} m_{2} = - 1 (⊥)

x_{v} = - \frac{b}{2 a}, y_{v} = f (x_{v})

Common misconceptions

"The perpendicular gradient to $m = 2$ is $- 2$ ." No: it is $- \frac{1}{2}$ (negative reciprocal).
"The vertex of $y = x^{2} - 4 x + 1$ is at $x = 4$ ." No: $x = - \frac{b}{2 a} = \frac{4}{2} = 2$ .
"Every parabola cuts the x-axis." No: if $Δ < 0$ it has no x-intercepts.

Worked examples

Example 1. Find the equation of the line through $(1, 3)$ and $(3, 7)$ .

Solution

$m = \frac{7 - 3}{3 - 1} = 2$ ; $y - 3 = 2 (x - 1)$ , so $y = 2 x + 1$ .

Example 2. Find the vertex and intercepts of $y = x^{2} - 4 x + 3$ .

Solution

$x_{v} = 2$ , $y_{v} = 4 - 8 + 3 = - 1$ : vertex $(2, - 1)$ . y-intercept $(0, 3)$ . $x^{2} - 4 x + 3 = (x - 1) (x - 3)$ : x-intercepts $(1, 0)$ and $(3, 0)$ .

Practice questions

Easier and harder questions are mixed, just like in a real exam.

Find the equation of the line through $(- 2, 3)$ and $(4, - 9)$ .
Show answer
$m = \frac{- 12}{6} = - 2$ ; $y - 3 = - 2 (x + 2)$ : $y = - 2 x - 1$ .
Common mistake: Using the gradient as the y-intercept.
Find the equation of the perpendicular bisector of $A (1, 2)$ and $B (5, 6)$ .
Show answer
Midpoint $(3, 4)$ ; $m_{A B} = 1$ , so $m = - 1$ : $y - 4 = - (x - 3)$ , $y = - x + 7$ .
Common mistake: Using the gradient of $A B$ itself.
Show that the lines $2 x - 3 y = 6$ and $3 x + 2 y = 1$ are perpendicular.
Show answer
Gradients $\frac{2}{3}$ and $- \frac{3}{2}$ ; product $- 1$ .
Common mistake: Reading the gradient from $2 x - 3 y = 6$ as 2.
Sketch $y = - x^{2} + 4$ showing intercepts and vertex.
Show answer
Opens down, vertex $(0, 4)$ , x-intercepts $(\pm 2, 0)$ , y-intercept $(0, 4)$ .
Common mistake: Drawing it opening upwards.
Complete a table of values for $y = x^{2} - 2 x - 3$ from $x = - 2$ to $x = 4$ .
Show answer
$x$ : $- 2, - 1, 0, 1, 2, 3, 4$ ; $y$ : $5, 0, - 3, - 4, - 3, 0, 5$ .
Common mistake: Computing $(- 2)^{2}$ as $- 4$ .
A company's profit is $P = - x^{2} + 60 x - 500$ (thousand FRW) for $x$ items. Find the number of items for maximum profit and the maximum profit.
Show answer
$x = 30$ ; $P = - 900 + 1800 - 500 = 400$ thousand FRW.
Common mistake: Answering with $x$ only.
Using the table, state the vertex, axis of symmetry, intercepts and the roots of $x^{2} - 2 x - 3 = 0$ .
Show answer
Vertex $(1, - 4)$ , axis $x = 1$ , y-intercept $(0, - 3)$ , roots $x = - 1$ and $x = 3$ .
Common mistake: Giving the y-intercept as a root.
Find the maximum value of $f (x) = - 2 x^{2} + 8 x - 3$ and where it occurs.
Show answer
$x = - \frac{8}{- 4} = 2$ ; $f (2) = - 8 + 16 - 3 = 5$ . Maximum 5 at $x = 2$ .
Common mistake: Sign error in $- \frac{b}{2 a}$ .

Matching and drag-and-drop

Matching

Match each item in column A with the correct answer in column B. Write the letter next to each number.

Column A	Column B
1. Find the vertex of $y = x^{2} - 6 x + 5$ .	a. $(3, - 4)$
2. Find the vertex of $y = 2 x^{2} - 8 x + 5$ .	b. 5
3. For $f (x) = 2 x^{2} - 3$ , find $f (- 2)$ .	c. $y = 2 x^{2} - 12 x + 10$
4. A parabola has x-intercepts 1 and 5 and passes through $(0, 10)$ . Find its equation.	d. $(2, - 3)$

Show answer

1 → a ( $(3, - 4)$ ) · 2 → d ( $(2, - 3)$ ) · 3 → b (5) · 4 → c ( $y = 2 x^{2} - 12 x + 10$ )

$f (3) = 9 - 18 + 5 = - 4$ .
Step 1. $x = - \frac{b}{2 a} = \frac{8}{4} = 2$ .
Step 2. $y = 2 (4) - 16 + 5 = - 3$ .
$2 \times 4 - 3$ .
Step 1. $y = a (x - 1) (x - 5)$ ; at $(0, 10)$ : $5 a = 10$ , so $a = 2$ .
Step 2. $y = 2 (x^{2} - 6 x + 5) = 2 x^{2} - 12 x + 10$ .

Drag and drop
Choose the correct word or phrase from the box for each gap. One word in the box is not needed.
4 · $y = 2 x^{2} - 4$ · $y = x^{2} - 2 x - 3$ · $- \frac{1}{4}$
1. The vertex of a parabola is $(1, - 4)$ and it passes through $(3, 0)$ . Its equation is .
2. A line parallel to $y = 4 x - 1$ has gradient .
3. A line perpendicular to $y = 4 x - 1$ has gradient .
Show answer
1: $y = x^{2} - 2 x - 3$ · 2: 4 · 3: $- \frac{1}{4}$
- Step 1. $y = a (x - 1)^{2} - 4$ .
  Step 2. $0 = 4 a - 4$ .
  Step 3. $a = 1$ : $x^{2} - 2 x + 1 - 4$ .
- Parallel lines have equal gradients.
- $m_{1} m_{2} = - 1$ .

Matching

Match each item in column A with the correct answer in column B. Write the letter next to each number.

Column A	Column B
1. The maximum value of $y = - x^{2} + 4 x$ is:	a. $y = - \frac{1}{2} x$
2. Find the points where $y = x^{2}$ meets $y = x + 6$ .	b. 20 m
3. Find the line through $(2, - 1)$ perpendicular to $y = 2 x + 3$ .	c. 4
4. A ball's height is $h = 20 t - 5 t^{2}$ metres after $t$ seconds. Find its maximum height.	d. $(3, 9)$ and $(- 2, 4)$

Show answer

1 → c (4) · 2 → d ( $(3, 9)$ and $(- 2, 4)$ ) · 3 → a ( $y = - \frac{1}{2} x$ ) · 4 → b (20 m)

Step 1. Vertex $x = 2$ .
Step 2. $y = - 4 + 8 = 4$ .
Step 1. $x^{2} = x + 6 \Rightarrow x^{2} - x - 6 = 0 \Rightarrow (x - 3) (x + 2) = 0$ .
Step 2. $x = 3$ gives $y = 9$ ; $x = - 2$ gives $y = 4$ .
Step 1. Perpendicular gradient: $- \frac{1}{2}$ (product $- 1$ ).
Step 2. $- 1 = - \frac{1}{2} (2) + c \Rightarrow c = 0$ , so $y = - \frac{1}{2} x$ .
Step 1. Maximum at $t = - \frac{20}{2 ( - 5 )} = 2$ s.
Step 2. $h = 40 - 20 = 20$ m.

Drag and drop
Choose the correct word or phrase from the box for each gap. One word in the box is not needed.
$(- 1, 0)$ and $(2, 3)$ · $(1, 2)$ and $(- 2, - 1)$ · $x = 2$ and $x = - 5$ · $y = 2 x + 1$
1. The line $y = x + 1$ and the parabola $y = x^{2} - 1$ meet at .
2. The line through $(1, 3)$ and $(3, 7)$ is .
3. The x-intercepts of $y = x^{2} + 3 x - 10$ are .
Show answer
1: $(- 1, 0)$ and $(2, 3)$ · 2: $y = 2 x + 1$ · 3: $x = 2$ and $x = - 5$
- Step 1. $x^{2} - 1 = x + 1$ .
  Step 2. $x^{2} - x - 2 = 0$ .
  Step 3. $x = - 1$ or 2.
- Step 1. $m = 2$ .
  Step 2. Then $y - 3 = 2 (x - 1)$ .
- $(x + 5) (x - 2) = 0$ .

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