Answer:
Option 1 - The vertex of the function is (h,k)=(-5,-28).
Step-by-step explanation:
Given : Function [tex]f(x)=x^2+10x-3[/tex]
To find : What are the coordinates of the vertex of the function?
Solution :
We have to convert the quadratic function [tex]y=ax^2+bx+c[/tex] into vertex form [tex]y=a(x-h)^2+k[/tex] where (h,k) are the vertex.
Function [tex]f(x)=x^2+10x-3[/tex]
Applying completing the square i.e. add and subtract half square of b,
[tex]f(x)=x^2+10x+5^2-5^2-3[/tex]
[tex]f(x)=(x+5)^2-25-3[/tex]
[tex]f(x)=(x+5)^2-28[/tex]
On comparing with vertex form,
h=-5 and k=-28
So, The vertex of the function is (h,k)=(-5,-28).
Therefore, Option 1 is correct.
