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

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Respuesta :

jimrgrant1 jimrgrant1 · Answer 1 · 2019-01-26T17:18:13+01:00

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

[tex]x_{vertex}[/tex] = - [tex]\frac{b}{2a}[/tex]

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

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

[tex]x_{vertex}[/tex] = - [tex]\frac{20}{10}[/tex] = - 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)

AlonsoDehner AlonsoDehner · Answer 2 · 2019-01-26T17:40:22+01:00

Answer:

(-2,-16)

Step-by-step explanation:

Given is a function

[tex]f(x) =5x^2+20x-16[/tex]

To find its vertex

We can use completion of squares method

[tex]f(x) =5x^2+20x-16[/tex]

[tex]f(x)=5(x^2+4x)-16\\=5(x^2+4x+4)-20-16\\=5(x+2)^2-16[/tex]

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