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

1 Answer
Jan 26, 2016

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

Explanation:

Starting with the intercepts:

The y-intercept is the value of #y# when #x=0#.
#y=(0+3)(0+5)=15#

The x-intercept(s) is/are the values of #x# when #y=0#
#0=(x+3)(x+5)#
#rArr x=-3# or #x=-5#

Determining the vertex.
Method 1:

For a parabola with a vertical axis,
the x coordinate of the vertex will be half way between the two x-intercepts.
i.e. the x-coordinate of the vertex will be #((-3)+(-5))/2 = -4#
By substituting #(-4)# for #x# in the given equation
#color(white)("XXX")y=(-4+3)(-4+5)=-1#
So the vertex will be at #(-4,-1)#

Method 2:

Convert the given equation into vertex form: #y=m(x-color(red)(a))^2+color(green)(b)#
for a parabola with vertex at #(color(red)(a),color(green)(b))#
#color(white)("XXX")y=(x+3)(x+5)#
#color(white)("XXX")rarr y=x^2+8x+15#
#color(white)("XXX")rarr y=x^2+8xcolor(blue)(+16) + 15 color(blue)(-16)#
#color(white)("XXX")rarr y=(x+4)^2+(-1)#
#color(white)("XXX")rarr y = (x- (color(red)(-4)))^2+(color(green)(-1))#