f(x)= -2x²-8x-5 a.)Express in vertex form b.) Find the zeros of the function c.) Find Y-intercept?

f(x) = -2x² - 8x - 5

a) 

-2x² - 8x - 5 = -2 * (x² + 4x + 5/2) = -2 * (x² + 4x + 4 - 4 + 5/2) = 

= -2 * (x² + 4x + 4 - 3/2) = -2 * [(x + 2)² - 3/2] = -2 * (x + 2)² + 3

Equation of f(x) in vertex form :

f(x) = -2 * (x + 2)² + 3 

b)

f(x) = 0 ===> -2 * (x + 2)² + 3 = 0 ===> -2 * (x + 2)² = -3 ===> (x + 2)² = 3/2 ===>

===> x + 2 = ±√(3/2) ===> x = -2 ± √(3/2) ===> x = -2 ± (√6)/2

Zeros of f(x) ===> (-2 - (√6)/2; 0) (-2 + (√6)/2; 0)

c)

f(0) = f(x = 0) = -2 * 0² - 8 * 0 - 5 = -5

y-intercept: (0; -5)

f(x) = -2x² - 8x - 5 => -2(x² + 4x) - 5

i.e. -2(x + 2)² - 5 + 8

so, f(x) = -2(x + 2)² + 3

The zeros occur when f(x) = 0

i.e. when  -2(x + 2)² + 3 = 0

so, (x + 2)² = 3/2

Then, x + 2 = ±√6/2

i.e. x = -2 ± √6/2

Hence, zeros at (-2 - √6/2, 0) and (-2 + √6/2, 0)

y-intercept at x = 0

so, f(0) = -2(2)² + 3 => -5

Hence, y-intercept at (0, -5)

:)>  

a. y - 3= -2(x + 2)²

b. x= -2 +/- √6 /2

c. (0, -5)