Question 60013
Hi Carol,
: 
For the quadratic function   y= f(x)= x^2-2x-15, find the vertex and x and y intercepts. Sketch the graph, and state the domain and range.
:
You can find the x coordinate of the vertex of a parabola when the quadratic equation is in the form of {{{y=f(x)=ax^2+bx+c}}}, with the formula {{{highlight(x=-b/2a)}}}
Your a=1, b=-2, and c=-15
{{{x=-(-2)/(2(1))}}}
{{{x=2/2}}}
{{{x=1}}}
You can find the y coordinate by letting x=1 and solving for y.
{{{y=(1)^2-2(1)-15}}}
{{{y=1-2-15}}}
{{{y=-16}}}
The vertex is (1,-16)
:
You can find the x-intercepts, if there are any, by letting y=0 and solve for x.  This time you can factor, sometimes you have to use the quadratic formula.
{{{0=x^2-2x-15}}}
0=(x-5)(x+3)
x-5=0 and x+3=0
x-5+5=0+5  and x+3-3=0-3
x=5 and x=-3
The x intercepts are (-3,0) and (5,0)
:
You can find the y-intercept by letting x=0:
{{{y=(0)^2-2(0)-15}}}
{{{y=0-0-15}}}
{{{y=-15}}}
The y-intercept is (0,-15)
The graph looks like this:
{{{graph(300,200,-10,10,-20,5,x^2-2x-15)}}}
Happy Calculating!!!