OMPT drill — Functions

Linear and quadratic functions

Slopes, intercepts, vertices, and intersections. If you can find where a line meets a parabola quickly, a surprising share of the test opens up. Attempt each problem before opening the solution; the notation only becomes automatic by writing it yourself.

Tested inOMPT-A OMPT-B OMPT-C OMPT-D OMPT-E OMPT-F OMPT-G

Lesson

A linear function $f (x) = a x + b$ draws a straight line; $a$ is the slope, how much $y$ changes when $x$ increases by one, and $b$ is where the line crosses the $y$ -axis. A quadratic function $f (x) = a x^{2} + b x + c$ draws a parabola, opening upward when $a > 0$ and downward when $a < 0$ , with everything interesting about it concentrated in one point: the vertex, sitting at $x = - \frac{b}{2 a}$ .

The OMPT asks about these two shapes constantly, but rarely in isolation. The productive question is where things meet: where a line crosses the axes, where two lines cross each other, where a line meets a parabola. Every one of those reduces to the same move, set two expressions equal and solve, so if your equation-solving from the algebra strand is in order, this topic is mostly about translating between picture and formula quickly.

Problem 1Numeric answer

A line passes through the points $(1, 3)$ and $(4, 12)$ . What is its slope?

Show the worked solution

Answer: 3

Slope is rise over run: $\frac{12 - 3}{4 - 1} = \frac{9}{3} = 3$ . I keep the subtraction order consistent — both differences go second point minus first point.

Problem 2Multiple choice

What is the vertex of the parabola $y = x^{2} - 6 x + 5$ ?

A $(3, - 4)$
B $(- 3, - 4)$
C $(3, 4)$
D $(6, 5)$

Show the worked solution

Answer: A — $(3, - 4)$

The vertex sits at $x = - \frac{b}{2 a} = - \frac{- 6}{2} = 3$ . Substituting back: $y = 9 - 18 + 5 = - 4$ . So the vertex is $(3, - 4)$ . Completing the square tells the same story: $y = (x - 3)^{2} - 4$ .

Problem 3Numeric answer

A line has slope $- 2$ and passes through $(3, 1)$ . What is its $y$ -intercept?

Show the worked solution

Answer: 7

I write $y = - 2 x + b$ and force the line through $(3, 1)$ : $1 = - 2 \cdot 3 + b$ , so $1 = - 6 + b$ and $b = 7$ . The line is $y = - 2 x + 7$ , which indeed hits $(0, 7)$ on the $y$ -axis.

Problem 4Multiple choice

The parabola $y = - 2 (x + 1)^{2} + 8$ crosses the $x$ -axis at two points. What is the sum of their $x$ -coordinates?

A $- 2$
B $2$
C $4$
D $- 3$

Show the worked solution

Answer: A — $- 2$

Setting $y = 0$ : $- 2 (x + 1)^{2} + 8 = 0$ gives $(x + 1)^{2} = 4$ , so $x + 1 = \pm 2$ and $x = 1$ or $x = - 3$ . The sum is $1 + (- 3) = - 2$ . Symmetry offers a shortcut: the intercepts straddle the axis of symmetry $x = - 1$ , so their sum must be $2 \cdot (- 1) = - 2$ .

Problem 5Spot the error

A student finds the $x$ -intercept of the line $y = 3 x - 12$ . Step 1: set $x = 0$ . Step 2: then $y = - 12$ . Step 3: so the $x$ -intercept is $(0, - 12)$ . Which step contains the error?

Show the worked solution

Answer: Step 1

The $x$ -intercept is where the line crosses the $x$ -axis, which means $y = 0$ , not $x = 0$ . Setting $y = 0$ : $0 = 3 x - 12$ , so $x = 4$ , and the intercept is $(4, 0)$ . What the student actually computed in Steps 2 and 3 is the $y$ -intercept, $(0, - 12)$ .

Problem 6Numeric answer

A ball is thrown upward and its height in metres after $t$ seconds is $h (t) = - 5 t^{2} + 20 t$ . After how many seconds does it reach its highest point?

Show the worked solution

Answer: 2

The height function is a downward parabola, so the maximum is at the vertex: $t = - \frac{b}{2 a} = - \frac{20}{2 \cdot ( - 5 )} = 2$ seconds. At that moment the height is $h (2) = - 20 + 40 = 20$ metres, but the question only asks for the time.