How do you find the vertex and the intercepts for #y=x^2+6x+5#?

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Answer 1 · 2017-08-07T14:42:50

Here is what I would do:

#y# intecept

When #x = 0#, we get #y = 5#

Vertex and #x# Intercepts

Note that the following method will not work if the equation has no #x# intercepts.

Also note that there are other ways to find the vertex.

#x# Intercepts

x^2+6x+5 = 0#

#(x+1)(x+5) = 0#

#(x+1) = 0# # " "# OR #" "# #(x+5) =0#

#x=-1# or #-5#

the #x# intercepts are #-1# and #-5#.

Vertex

The vertex is midway between the #x# intercepts. So it is at the average of the #x# intercepts.

The average is #1/2# of the sum, so the vertex is at

#x = 1/2((1-)+(-5)) = (-6)/2 = -3#

To find the #y# coordinate of the vertex, plug #-3# in for #x# in the equation

#y = x^2+6x+5# at #x = -3# is

#(-3)^2+6(-3)+5 = 9-18+5 = -9+5=-4#

The vertex is #(-3,-4)#