Question

Identify the axis of symmetry and vertex of f(x) = –2x2 –2x–1.

Axis of symmetry: x = – 0.5; Vertex: (– 0.5, – 0.50)
Axis of symmetry: x = 0.5; Vertex: (– 0.5, – 0.50)
Axis of symmetry: x = 0.5; Vertex: (0.5, – 0.50)
Axis of symmetry: x = – 0.5; Vertex: (– 0.5, 0.50)

Answer 1

A. The vertex is at (-0.5, -0.5) and the axis of symmetry is x = -0.5.

Explanation:

–2x2 –2x–1

Convert to vertex form:

= -2(x^2 + x) - 1

= -2 [ (x + 0.5x)^2- 0.25] - 1

= -2(x + 0.5)^2 + 0.5 - 1

= -2(x + 0.5)^2 - 0.5.

Compare this with the general vertex form:

a(x - b)^2 + c where the vertex is at (b, c) and the axis of symmetry is

x = b

we see that b = -0.5 and c = -0.5 therefore

the vertex is at (-0.5, -0.5) and the axis of symmetry is x = -0.5.

Answer 2

The first option

Explanation:

Given a quadratic function in standard form : ax² + bx + c : a ≠ 0, then

The equation of the axis of symmetry which is also the x- coordinate of the vertex is found using

x = -
`(b)/(2a)`

f(x) = - 2x² - 2x - 1 ← is in standard form

with a = - 2, b = - 2, so

x = -
`(-2)/(-4)` = - 0.5

Equation of axis of symmetry is x = - 0.5

Substitute x = - 0.5 into f(x)

f(- 0.5) = - 2(- 0.5)² - 2(- 0.5) - 1 = - 0.5 + 1 - 1 = - 0.5

Vertex = (- 0.5, - 0.5 )