Quadratic function: forms and graph practice questions.

State the equation of the axis of symmetry of the graph of y = 2x² + 8x − 3.

• x = 2
• x = 4
• x = −2
• x = −4

Answer: x = −2

The axis of symmetry is x = −b/(2a) = −8/(2 × 2) = −2. Forgetting the minus sign or the factor 2a gives the distractors.

A quadratic is given in factored form as y = (x + 3)(x − 5). Its axis of symmetry is

• x = 1
• x = −1
• x = 2
• x = −15

Answer: x = 1

The x-intercepts are x = −3 and x = 5, and the axis of symmetry lies midway between them: x = (−3 + 5)/2 = 1.

Determine the y-intercept of the graph of y = 3(x − 1)² − 5.

• −5
• −2
• 3
• −8

Answer: −2

Substitute x = 0: y = 3(−1)² − 5 = 3 − 5 = −2. Reading off −5 confuses the constant k in vertex form with the y-intercept.

The graph of y = 2(x − 1)(x + 3) crosses the x-axis at

• x = −1 and x = 3
• x = 2 and x = −6
• x = 1 and x = 3
• x = 1 and x = −3

Answer: x = 1 and x = −3

Each factor is zero when x = 1 or x = −3; note the sign flip from (x − 1) to root x = +1. The leading factor 2 does not move the x-intercepts.

The curve y = −2(x − 3)² + 8 has a maximum point at

• (3, 8)
• (−3, 8)
• (3, −8)
• (8, 3)

Answer: (3, 8)

In vertex form y = a(x − h)² + k the vertex is (h, k) = (3, 8); since a = −2 < 0 the parabola opens downward, so this vertex is a maximum.

A parabola has x-intercepts at x = −1 and x = 4, with leading coefficient 1. Its equation is

• y = x² + 3x − 4
• y = x² − 5x + 4
• y = x² − 3x − 4
• y = x² + 5x − 4

Answer: y = x² − 3x − 4

y = (x + 1)(x − 4) = x² − 3x − 4. Expanding with sign errors gives x² + 3x − 4; the intercepts +1 and +4 would give x² − 5x + 4.