Question

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

Answer 1

the vertex is (-2,-36)

Answer 2

Vertex of given quadratic function
`f(x)=5x^2+20x-16` is (-2,-36)

Explanation:

Given function
`f(x)=5x^2+20x-16`

We have to find the vertex of the given function
`f(x)=5x^2+20x-16`.

The standard quadratic function is represented by
`f(x)=a(x-h)^2+k` ,

Where (h,k) represents the vertex and a ≠ 0.

If a is positive, the graph opens upward, and if a is negative, then it opens downward.

We first write the given equation in standard form ,by using completing square,

We know
`(a-b)^2=a^2-2ab+b^2`

First taking 5 common from function, we have,

`f(x)=5(x^2+4x-(16)/(5))`

Comparing we have, a = x and

-2ab= +4x ⇒ -2b = 4 ⇒ b = -2

Add and subtract b
`b^2=4` , we get,

`f(x)=5(x^2+4x+4-4-(16)/(5))`

On simplifying, we have,

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

`f(x)=5((x+2)^2+(-36)/(5))`

We get,
`f(x)=5((x+2)^2)-36`

On comapring with standard equtaion , we have h = -2 and k = -36

Thus, vertex of given quadratic function
`f(x)=5x^2+20x-16` is (-2 ,-36)