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

Answer:

(-5,-28)

Step-by-step 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;

$x = \frac{-b}{2a}$

Substitute 10 for b and 1 for a

$x = \frac{-10}{2*1}$

$x = \frac{-10}{2}$

$x = -5$

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)