Question

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

Answer 1

Important points:
`color(white)("XXX")`x-intercepts
`color(white)("XXX")`y-intercept
`color(white)("XXX")`vertex

Explanation:

The x-intercepts
These are the values of `x` when `y` (or in this case `f(x)`) `=0`
`color(white)("XXX")f(x)=0`
`color(white)("XXX")rarr (x+2)=0 or (x-5)=0`
`color(white)("XXX")rarr x=-2 or x=5`
So the x-intercepts are at `(-2,0)` and `(5,0)`

The y-intercept
This is the value of `y` (`f(x)`) when `x=0`
`color(white)("XXX")f(x)=(0+2)(0-5)=-10`
So the y(`f(x)`)-intercept is at `(0,-10)`

The vertex
There are several ways to find this;
I will use conversion to vertex form `f(x)=(x-color(red)(a))^2+color(blue)(b)` with vertex at `(color(red)(a),color(blue)(b))`
`color(white)("XXX")f(x)=(x+2)(x-5)`
`color(white)("XXX")rarr f(x)=x^2-3x-10`
`color(white)("XXX")rarr f(x)=x^2-3xcolor(green)(+(3/2)^2) -10 color(green)(-(3/2)^2)`
`color(white)("XXX")rarr f(x)=(x-color(red)(3/2))^2+(color(blue)(-49/4))`
So the vertex is at `(3/2,-49/4)`

Here is what the graph should look like:
graph{(y-(x+2)(x-5))(x^2+(y+10)^2-0.05)((x+2)^2+y^2-0.05)((x-5)^2+y^2-0.05)((x-3/2)^2+(y+49/4)^2-0.05)=0 [-14.52, 13.96, -13.24, 1.01]}