Need Algebra 2 quadratic help?

Question

f(x)= 2(x+3)^2 +5 

Find the vertex and x intercepts. 

Show steps, plz. 
Thnx!

Answer 1

(a).The equation is y = 2(x + 3)² + 5.

The above equation represent the parabola.

Compare the equation y = 2(x + 3)² + 5 with standard form of the equation of a parabola with vertex (h, k) and axis of symmetry x = h is y = a(x - h)² + k.

Vertex (h, k ) = (-3, 5), and axis of symmetry x = h = -3.

To find the x - intercept, substitute y = 0 in the y = 2(x + 3)² + 5.

0 = 2(x + 3)² + 5.

2(x + 3)² = - 5

(x + 3)² = - 5/2

x + 3 = √ (- 5/2)

x = ±i√ (5/2) - 3

Therefore x intercepts are -3 + i√ (5/2) and -3 - i√ (5/2).

Vertex (h, k ) = (-3, 5) and x intercepts are -3 + i√ (5/2) and -3 - i√ (5/2).