What is the vertex form of #y= y=x^2+5x-36#?

2 Answers

The vertex form #y--169/4=(x--5/2)^2#
with vertex at #(h, k)=(-5/2, -169/4)#

Explanation:

From the given equation #y=x^2+5x-36#

complete the square

#y=x^2+5x-36#
#y=x^2+5x+25/4-25/4-36#

We group the first three terms

#y=(x^2+5x+25/4)-25/4-36#

#y=(x+5/2)^2-25/4-144/4#

#y=(x+5/2)^2-169/4#

#y--169/4=(x--5/2)^2#

graph{y+169/4=(x--5/2)^2[-100, 100,-50,50]}

God bless...I hope the explanation is useful.

Mar 31, 2016

#y = (x + 5/2)^2 - 169/4#

Explanation:

x-coordinate of vertex:
#x = -b/(2a) = -5/2#
y-coordinate of vertex:
#y(-5/2) = (25/4) - 25/2 - 36 = -25/4 - 36 = -169/4.#
#Vertex (-5/2, - 169/4)#
Vertex form: #y = (x + 5/2)^2 - 169/4#