Question 137128
You can find the x- and y-intercepts of a quadratic function {{{y = ax^2+bx+c}}}as follows:
1) Set x = 0 and solve for y. This will give you the y-intercept.
2) Set y = 0 and solve for x. This will give you the x-intercept.
Here's an example:
{{{y = x^2-8x+12}}} Find the x- and y-intercepts. 
1) Set x = 0 and solve for y.
{{{y = (0)^2-8(0)+12}}}
{{{y = 12}}} The y-intercept is: (0, 12)
2) Set y = 0 and solve for x.
{{{0 = x^2-8x+12}}} Factor the trinomial:
{{{(x-2)(x-6) = 0}}} from which we get:
{{{x = 2}}} and {{{x = 6}}} These are the x-intercepts. These are also called the "zeros" or the "roots" of the equation and notice that they are "real" numbers, which really means that the equation has real x-intercepts.
Of course, not all quadratic equations will have x-intercepts but they will alway have a y-intercept.
Here's a graph of the given example:
{{{graph(400,400,-5,8,-5,15,x^2-8x+12)}}}
Here's an example in which the quadratic equation has no x-intercepts.
{{{y = x^2+4x+5}}}
1) Set x = 0 and solve for y.
{{{y = (0)^2+4(0)+5}}}
{{{y = 5}}} This is the y-intercept.
2) Set y = 0 and solve for x.
{{{0 = x^2+4x+5}}} Use the quadratic formula to solve: (You can do the details)
{{{x = -2+i}}} and {{{x = -2-i}}} No "real" roots so no x-intercepts.
Here's the graph:
{{{graph(400,400,-5,5,-5,8,x^2+4x+5)}}}