Question

consider the function f(x)=15x^2+60-19 Part A: Write the function in vertex form. Part B: Name the vertex for the function

Answer 1

Part A) The function written in vertex form is
`f(x)=15(x+2)^(2)-79`

Part B) The vertex of the function is the point
`(-2,-79)`

Explanation:

Part A) Write the function in vertex form

we know that

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

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

where

(h,k) is the vertex of the parabola

In this problem we have

`f(x)=15x^(2)+60x-19`

Convert to vertex form

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

`f(x)+19=15x^(2)+60x`

Factor the leading coefficient

`f(x)+19=15(x^(2)+4x)`

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

`f(x)+19+60=15(x^(2)+4x+4)`

`f(x)+79=15(x^(2)+4x+4)`

Rewrite as perfect squares

`f(x)+79=15(x+2)^(2)`

`f(x)=15(x+2)^(2)-79` -----> function in vertex form

Part B) Name the vertex for the function

we have

`f(x)=15(x+2)^(2)-79`

The vertex of the function is the point
`(-2,-79)`

The parabola open upward, so the vertex is a minimum

see the attached figure to better understand the problem