SOLUTION: How do I compute the x and y axis intercepts of a quadratic function? Please provide an example to illustrate the explanation. Thank you

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Question 137128: How do I compute the x and y axis intercepts of a quadratic function? Please provide an example to illustrate the explanation. Thank you
Found 2 solutions by oscargut, Earlsdon:
Answer by oscargut(2103) About Me  (Show Source):
You can put this solution on YOUR website!
if f(x)=ax^2+bx+c
then
y axis intercept is P=(0,f(0))=(0,c)
x axis intercept are the roots of f(x)
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
f(x)=x^2-5x+6
y axis intercept is P=(0,6)
x+=+%285+%2B-+sqrt%28+25-4%2A1%2A6+%29%29%2F%282%2A1%29+
x axis intercept are x=2 and x=3
point are (2,0) and (3,0)

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
You can find the x- and y-intercepts of a quadratic function y+=+ax%5E2%2Bbx%2Bcas 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%5E2-8x%2B12 Find the x- and y-intercepts.
1) Set x = 0 and solve for y.
y+=+%280%29%5E2-8%280%29%2B12
y+=+12 The y-intercept is: (0, 12)
2) Set y = 0 and solve for x.
0+=+x%5E2-8x%2B12 Factor the trinomial:
%28x-2%29%28x-6%29+=+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%28400%2C400%2C-5%2C8%2C-5%2C15%2Cx%5E2-8x%2B12%29
Here's an example in which the quadratic equation has no x-intercepts.
y+=+x%5E2%2B4x%2B5
1) Set x = 0 and solve for y.
y+=+%280%29%5E2%2B4%280%29%2B5
y+=+5 This is the y-intercept.
2) Set y = 0 and solve for x.
0+=+x%5E2%2B4x%2B5 Use the quadratic formula to solve: (You can do the details)
x+=+-2%2Bi and x+=+-2-i No "real" roots so no x-intercepts.
Here's the graph:
graph%28400%2C400%2C-5%2C5%2C-5%2C8%2Cx%5E2%2B4x%2B5%29