Answer:
(-5,-28)
Step-by-step explanation:
Given
[tex]f(x)=x^2+10x - 3[/tex]
Required
Find the coordinates of the vertex
Given that the function is a quadratic function;
The general form of a quadratic function is
[tex]f(x)= ax^2+bx +c[/tex]
By comparison;
[tex]a = 1\ b = 10\ and\ c = -10[/tex]
To calculate the coordinates of the vertex;
we start by calculating the x-coordinate;
[tex]x = \frac{-b}{2a}[/tex]
Substitute 10 for b and 1 for a
[tex]x = \frac{-10}{2*1}[/tex]
[tex]x = \frac{-10}{2}[/tex]
[tex]x = -5[/tex]
Substitute [tex]x = -5[/tex] in the given function;
[tex]y=x^2+10x - 3[/tex]
[tex]y=(-5)^2+10*-5 - 3[/tex]
[tex]y=25-50 - 3[/tex]
[tex]y=-28[/tex]
Hence, the coordinates of the vertex is (-5,-28)
