Question

What are the coordinates of the vertex of the function f(x)=x^2+10x−3? (–5, –28) (–5, 28) (5, –28) (5, 28)

Answer 1

(-5,-28)

Explanation:

Given

`f(x)=x^2+10x - 3`

Required

Find the coordinates of the vertex

Given that the function is a quadratic function;

The general form of a quadratic function is

`f(x)= ax^2+bx +c`

By comparison;

`a = 1\ b = 10\ and\ c = -10`

To calculate the coordinates of the vertex;

we start by calculating the x-coordinate;

`x = (-b)/(2a)`

Substitute 10 for b and 1 for a

`x = (-10)/(2*1)`

`x = (-10)/(2)`

`x = -5`

Substitute
`x = -5` in the given function;

`y=x^2+10x - 3`

`y=(-5)^2+10*-5 - 3`

`y=25-50 - 3`

`y=-28`

Hence, the coordinates of the vertex is (-5,-28)