What are the important points needed to graph #f(x) = (x-2)(x + 5)#?

1 Answer
Jun 16, 2018

x-intercepts
#x=-5,x=2#
y-intercept
#y=-10#
vertex: #(-3/2,-49/4)#

Explanation:

You are given the x-intercepts

#(x-2)(x+5)#
#x=2#
#x=-5#

First find y-intercept by multiplying out to standard form #Ax^2+Bx+C# and set x to 0

#f(x)=(x-2)(x+5)=x^2+3x-10#
#f(x)=(0)^2+3(0)-10=-10#

y-intercept is at #y=-10#

Next convert to vertex form by completing the square

#x^2+3x=10#

Divide coefficient by 2 and square

#(3/2)^2 = 9/4#
#(x^2+3x + 9/4)=10+9/4#

Rewrite

#(x+3/2)^2=40/4 + 9/4=49/4#
#f(x)=(x+3/2)^2-49/4#

Vertex is #(-3/2, -49/4)# or #(-1.5, -12.25)#

graph{(x+3/2)^2-49/4 [-21.67, 18.33, -14.08, 5.92]}