Question

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

Answer 1

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]}