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

1 Answer
michael
Jul 26, 2017

Generally, how to find...
I. X-intercept(s)
- set y=0 and solve for x.
II. Y-intercept(s)
- set x=0 and solve for y.
III. Vertex
i. FOIL (distribute)
ii. vertex: (-b/2a, f(-b/2a))

For this question...
I. X-intercepts
#y=(x+5)(x+3)#
#0=(x+5)(x+3)#
#x=-5, -3#
x-intercepts: (-5,0), (-3,0)

II. Y-intercept
#y=(x+5)(x+3)#
#y=(0+5)(0+3)#
#y=15#
y-intercept: (0,15)

III. Vertex
#y=(x+5)(x+3)#
#y=x^2+8x+15#
#-b/(2a)=-((8))/(2*(1))=-4#
This is the x-value. To find the y-value plug in -4 for x and solve for y.
#y=(-4+5)(-4+3)#
y=-1
The vertex is (-4,-1)

Related questions
See all questions in Vertex Form of a Quadratic Equation
Impact of this question
1722 views around the world
You can reuse this answer
Creative Commons License