How do you find the vertex and intercepts for #y=(x+5)(x+3)#?

1 Answer
Dylan D.
Apr 21, 2018

the #x#-intercepts are #-5# and #-3#, the y-intercept is #15# and the vertex is #(-4,-1)#

Explanation:

Because the equation is in the form #y=a(x-p)(x-q)#, we know the #x#-intercepts are #p# and #q#
Therefore, the #x#-intercepts are at #-5# and #-3#
Note the negative signs and the fact that in the case #a=1# so it is committed.
This makes sense because the #x#-intercepts are whenever #y=0#. This occurs when either bracket is #0#.

The #x#-value of the vertex is the midvalue of the intercepts.
let #V_x# be the #x#-value of the vertex
#V_x=((-5)+(-3))/2#
#V_x=-4#
Subbing in #x=-4# to solve for the #y#-value of the vertex:
Let this be #V_y#:
#V_y=((-4)+5)((-4)+3)#
#V_y =(1)(-1)#
#V_y = -1#

Therefore, the vertex is #(-4,-1)#

To find the #y#-intercept we substitute #x=0# into the equation.
#y=(5)(3)#
#y=15#

In conclusion, the #x#-intercepts are #-5# and #-3#, the #y#-intercept is #15# and the vertex is #(-4,-1)#

Included below are graphs of the equation:
graph{(x+5)(x+3) [-7.757, 3.6, -2.454, 3.224]}
graph{(x+5)(x+3) [-49.16, 45, -23.16, 23.9]}

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