Question 65159
f(x) = x^2 - 6x
There are a couple of ways to find the vertex, if your teacher is putting the equation in vertex form, let me know and I'll redo this.
When a quadratic equation is in this form: f(x)=ax^2+bx+c, I prefer using this formula to find the x coordinate of the vertex:  {{{highlight(x=-b/2a)}}}
a=1, b=-6 and c=0
{{{x=-(-6)/(2(1))}}}
{{{x=6/2}}}
{{{x=3}}}
To find the y-coordinate find f(3)
{{{f(3)=(3)^2-6(3)}}}
{{{f(3)=9-18}}}
{{{f(3)=-9}}}
The vertex (x,y)=(3,-9)
You find the y-intercept, by letting x=0:
{{{f(0)=(0)^2-6(0)}}}
{{{f(0)=0-0}}}
{{{f(0)=0}}}
The y-intercept is (0,0)  The origin will be both an x and a y-intercept.
To find the x-intercept, let f(x)=0 and solve for x.
{{{0=x^2-6x}}}
{{{0=x(x-6)}}}
x=0 and x-6=0
x=0 and x=6
The x-intercepts are (0,0) and (6,0)
Here's what it looks like:
{{{graph(300,200,-10,10,-10,10,x^2-6x)}}}
Happy Calculating!!!