Question

Find the vertex and the y-intercept of the graph of y = 2x^2 + 16x + 33.

Answer 1

Answer:

see explanation

Step-by-step explanation:

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

Then the x- coordinate of the vertex is

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

y = 2x² + 16x + 33 ← is in standard form

with a = 2, b = 16, c = 33, hence

$x_{vertex}$ = - $\frac{16}{4}$ = - 4

Substitute x = - 4 into the equation for y- coordinate of vertex

y = 2(- 4)² + 16(- 4) + 33 = 32 - 64 + 33 = 1

vertex = (- 4, 1 )

To find the y- intercept substitute x = 0 into the equation

y- intercept = 2(0)² + 16(0) + 33 = 0 + 0 + 33 = 33