The math equation is f(x)=(x+1)(x+5). How do I find the vertex of this without graphing it?

Question

(x-5)(x+1)

Answer 1

f(x) = (x+1)(x+5) has x intercepts = (-1,0) and (-5,0). vertex has an x coordinate half way between -1 and 5 = (-1-5)/2 = -6/2 = -3

= x^2 +6x +5

complete the square, take half the x coefficient and square it, to get (6/2)^2 = 3^2 = 9. add & subtract it.

= x^2 +6x + 9 + 5- 9

= (x+3)^2 -4 in vertex form = (x+h)^2 -k where (h,k) = the vertex = (-3,-4)

Answer 2

Hi Catarina C

Another quick way is to simply use the FOIL method on the function

f(x)=(x+1)(x+5)=x²+6x+5 now it’s in the form of ax²+bx+c the Quadratic Equation

and for your function:

a=1, b=6 and c=5

In this form the x coordinate of the vertex is -b/2a, for your function -6/2= -3

The y coordinate of the vertex is found by finding f(-3)

f(-3) = (-3)²+ 6(-3)+5 = 9 - 18 + 5 = -18 + 14 = -4

Your vertex is (-3,-4)