How do you find the vertex and the intercepts for #y= (x-4) (x+2)#?

1 Answer
Patrick H.
Jun 26, 2018

See below

Explanation:

For the vertex, we need to get our equation in standard form. Using FOIL, it can be rewritten as #y=x^2-2x-8#. To find its x-coordinate, use the formula #x=-b/(2a)#:

#x=(-(-2))/(2(1))=2/2=1#

The y-coordinate is found by plugging our answer back into the equation:

#y=(1)^2-2(1)-8#
#y=1-2-8=-7#

Vertex: #(1,-7)#

By "intercepts", I believe you mean both the x-intercepts and the y-intercept. For the x-intercepts, we can use the factored equation, set #y# to #0#, and solve:

#0=(x-4)(x+2)#
#x-4=0#
#x=4#
#x+2=0#
#x=-2#

X-intercepts: #(4,0), (-2,0)#

To find the y-intercept, we will use the standard form of our equation, set #x# to #0#, and solve:

#y=0^2-2(0)-8#
#y=-8#

Y-intercept: #(0,-8)#

Hope this helped!

Related questions
See all questions in Vertex Form of a Quadratic Equation
Impact of this question
4278 views around the world
You can reuse this answer
Creative Commons License