Question 64855
f(x)=x^2+6x+5
Find the x value of the vertex (turning point)
There are 2 ways to do this let me know if this isn't the way your teacher is doing it.
Because this is in f(x)=ax^2+bx+c form, I would use the formula {{{highlight(x=-b/2a)}}} to find the x-coordinate, then find f(-b/2a) to find the y-coordinate.
a=1, b=6
{{{x=-(6)/(2(1))}}}
{{{x=-6/2}}}
x=-3
y={{{f(-3)=(-3)^2+6(-3)+5}}}
{{{y=9-18+5}}}
y=-4
The vertex is (-3,-4)
Happy Calculating!!!