Question

What is the vertex of f (x) = 5x^2+20x-16

Answer 1

Answer:

(-2,-16)

Step-by-step explanation:

Given is a function

$f(x) =5x^2+20x-16$

To find its vertex

We can use completion of squares method

$f(x) =5x^2+20x-16$

$f(x)=5(x^2+4x)-16\\=5(x^2+4x+4)-20-16\\=5(x+2)^2-16$

This is in std vertex form of a parabola

From the equation we find that the vertex is

(-2,-16)

Hence vertex is (-2,-16)

Verify:

Derivative of f(x) =f'(x)=10x+20

Equate to 0 to have x=-2

f(-2) = 20-40-16=-36

Thus verified

Answer 2

Answer:

vertex = (- 2, - 36)

Step-by-step explanation:

Given a parabola in standard form : y = ax² + bx + c : a ≠ 0

Then the x- coordinate of the vertex is

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

f(x) = 5x² + 20x - 16 is in standard form

with a = 5, b = 20 and c = - 16

$x_{vertex}$ = - $\frac{20}{10}$ = - 2

Substitute x = - 2 into f(x) for corresponding y- coordinate

f(- 2) = 5(- 2)² +20(- 2) - 16 = 20 - 40 - 16 = - 36

vertex = (- 2, - 36)