Question 64803
Find the vertex and y-intercept of:
{{{f(x) = x^2 - 8x - 20}}} Rewrite as:
{{{y = x^2 - 8x - 20}}}
The x-coordinate of the vertex is given by:
{{{-b/2a}}} (See the general form:{{{ax^2 + bx + c = 0}}}
In your equation, a = 1 and b = -8
{{{x = -(-8)/2(1)}}}
{{{x = 4}}} Substitute this into the original equation and solve for y.
{{{y = (4)^2 - 8(4) - 20}}}
{{{y = 16 - 32 - 20}}}
{{{y = -36}}} 
The vertex is at: (4, -36)

The y-intercept is found by substituting x = 0 and solving for y.
{{{y = (0)^2 - 8(0) - 20}}}
{{{y = -20}}}
The y-intercept is at: (0, -20)
{{{graph(300,200,-5,12,-38,5,x^2-8x-20)}}}