SOLUTION: f(x)=x^2+6x+5 Find the x value of the vertex (turning point)

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: f(x)=x^2+6x+5 Find the x value of the vertex (turning point)      Log On


   

Question 64855: f(x)=x^2+6x+5
Find the x value of the vertex (turning point)
Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
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%28x=-b%2F2a%29 to find the x-coordinate, then find f(-b/2a) to find the y-coordinate.
a=1, b=6
x=-%286%29%2F%282%281%29%29
x=-6%2F2
x=-3
y=f%28-3%29=%28-3%29%5E2%2B6%28-3%29%2B5
y=9-18%2B5
y=-4
The vertex is (-3,-4)
Happy Calculating!!!