Question

(Easy) For the following quadratic function, find the axis of symmetry, the vertex and the y-intercept. y = x^2 + 12x + 32

Answer 1

Answer:

$y ( \times + 6) 2 - 4$

Answer 2

Answer:

see explanation

Step-by-step explanation:

Given a quadratic in standard form , y = ax² + bx + c ( a ≠ 0 ), then

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

$x_{vertex}$ = - $\frac{b}{2a}$

y = x² + 12x + 32 ← is in standard form

with a = 1, b = 12 , then

$x_{vertex}$ = - $\frac{12}{2}$ = - 6

Substitute x = - 6 into y for corresponding y- coordinate

y = (- 6)² + 12(- 6) + 32 = 36 - 72 + 32 = - 4

Thus

equation of axis of symmetry is x = - 6

vertex = (- 6, - 4 )

To find the y- intercept , let x = 0

y = 0² + 12(0) + 32 = 32 ← y- intercept