SOLUTION: Please solve this quadratic to standard form equation http://i.imgur.com/foBgfZS.png

Question 904411: Please solve this quadratic to standard form equation
http://i.imgur.com/foBgfZS.png
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
quadratic equation is f(x) = x^2 + 10x + 24

standard form, or vertex form is f(x) = a(x-h)^2 + k

to convert, complete the squares to get:

f(x) = (x+5)^2 -25 + 24 which becomes f(x) = (x+5)^2 - 1.

if you need a lesson on how to complete the squares, you can find it here.
http://www.mathsisfun.com/algebra/completing-square.html

the vertex is (-5,-1)

to find the x intercepts, set f(x) equal to 0 and solve for x.

you get (x+5)^2 - 1 = 0
add 1 to both sides of the equation to get (x+5)^2 = 1.
take the square root of both sides of the equation to get x+5 = +/- sqrt(1).
subtract 5 from both sides of the equation to get x = -5 +/- sqrt(1).
simplify to get x = -5 +/- 1
solve for x to get
x = -6 or x = -4.
the coordinates of the x intercepts are (-6,0) and (-4,0).

to find the y intercept, set x = 0 and solve for f(x).

you get f(0) = (0+5)^2 - 1 which becomes f(0) = 5^2 - 1 which becomes f(0) = 25 - 1 which becomes f(0) = 24.

your y intercept is y = 24
the coordinate of the y intercept is (0,24)

your solutions are:

vertex = (-5,-1)
x intercepts are (-6,0) and (-4,0)
y intercept is (0,24)

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