Question

What is the vertex of the graph y=x^2+4x-1

a. (1,4)
b. (0,-1)
c. (-1,-4)
d. (-2,-5)

Answer 1

Option d. (-2,-5)

Explanation:

we know that

The equation of a vertical parabola in vertex form is equal to

`y=a(x-h)^(2) +k`

where

a is a coefficient

(h,k) is the vertex of the parabola

In this problem we have

`y=x^(2) +4x-1`

This is the equation of a vertical parabola open up

The vertex is minimum

Convert the equation in vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`y+1=x^(2) +4x`

Complete the square. Remember to balance the equation by adding the same constants to each side.

`y+1+4=x^(2) +4x+4`

`y+5=x^(2)+4x+4`

Rewrite as perfect squares

`y+5=(x+2)^(2)`

`y=(x+2)^(2)-5` -----> equation in vertex form

therefore

The vertex is (-2,-5)