Functions and Graphs Practice Questions for Matric

Practice questions with solutions

Find the turning point of y = x² − 4x + 1

1. x = −b/2a = 2
2. y = 4 − 8 + 1 = −3

Answer: (2, −3)

State the domain of f(x) = 1/(x − 3)

1. Denominator ≠ 0
2. x ≠ 3

Answer: x ∈ ℝ, x ≠ 3

Find x-intercepts of y = x² − 9

1. Set y = 0
2. x² = 9

Answer: x = ±3

Find the asymptotes of y = 3/(x + 2) − 1

1. Vertical asymptote: x + 2 = 0, so x = −2
2. Horizontal asymptote: y = q = −1

Answer: x = −2 and y = −1

Find the inverse of f(x) = 2x − 6

1. Write y = 2x − 6
2. Swap x and y: x = 2y − 6
3. Make y the subject: y = (x + 6)/2

Answer: f⁻¹(x) = x/2 + 3

An exponential graph y = bˣ + 2 passes through (1; 5). Find b.

1. Substitute the point: 5 = b¹ + 2
2. b = 3

Answer: y = 3ˣ + 2

For which values of x is y = x² − 4x + 1 decreasing?

1. The axis of symmetry is x = −b/2a = 2
2. The parabola opens upwards, so it decreases to the left of the turning point

Answer: x < 2

Find the range of y = −2(x − 1)² + 5

1. a = −2 < 0, so the parabola has a maximum
2. The turning point is (1; 5)

Answer: y ≤ 5

Find the inverse of g(x) = 2ˣ and state its domain.

1. Swap x and y: x = 2ʸ
2. Write y as the subject: y = log₂ x

Answer: g⁻¹(x) = log₂ x, with x > 0

A parabola has turning point (1; −4) and passes through (3; 0). Find its equation.

1. Use y = a(x − p)² + q: y = a(x − 1)² − 4
2. Substitute (3; 0): 0 = 4a − 4, so a = 1

Answer: y = (x − 1)² − 4, or y = x² − 2x − 3

Find the average gradient of f(x) = x² between x = 1 and x = 3.

1. f(1) = 1 and f(3) = 9
2. Average gradient = (9 − 1) / (3 − 1)

Answer: 4

Tips

• Identify the function family before sketching.
• Always plot intercepts and asymptotes first.
• Draw asymptotes as dotted lines and label every intercept and turning point.
• Check your sketch against one extra point substituted into the equation.