What is the vertex form of #y=x^2 + 8x - 9#?

1 Answer
Dec 30, 2015

#y=(x+4)^2 - 25# The vertex form is obtained by completing the square. Please go through the explanation on how we can get the solution.

Explanation:

To find the vertex form of the quadratic function we need to make the function into #y=a(x-h)^2+k# where #(h,k)# is the vertex.

#y=x^2+8x - 9#

Knowledge about completing squares would help.

#y+9 =x^2 + 8x# Firstly move the constant to the other side, this can be done by adding 9 to both the sides.

We need to make #x^2+8x# into a whole square.

*In case you are confused with what a whole square is, it is nothing but expressing the given expression in the form #(x-h)^2#

On expanding #(x-h)^2# would be #x^2-2hx + h^2#

Now let us see that the coefficient of #x# is #2h# and we need #h^2# so we divide coefficient of #x# by #2# and square the result to get #h^2# That would help completing the square. The steps to complete the square for our question is given below.*

Divide the coefficient of #x# by #2# and square the result and add it to both the sides.

#y+9 +(8/2)^2 = x^2 + 8x + (8/2)^2#
#y+9+16 = x^2 + 8x + 16#

#y+25= (x+4)^2#

#y=(x+4)^2 - 25# Vertex form