Expert: Sherman D. - 12/5/2005

Question
Why is the axis located at -2a? Why the 2?
Someone once told me it's related to the formula for completing the square, the doubling of the middle term, but how and why?
Rachim

Answer
ax^2 + bx + c = 0
ax^2 + bx = -c
x^2 + (b/a)x = (-c/a)
x^2 + (b/a)x + ((b/a)/2)^2 = (-c/a) + ((b/a)/2)^2
x^2 + (b/a)x + (b/(2a))^2 = (b/(2a))^2 - (c/a)
x^2 + (b/a)x + ((b^2)/(4a^2)) = ((b^2)/(4a^2)) - (c/a)

factor the left side into a perfect square

(x + (b/(2a)))^2 = ((b^2)/(4a^2)) - (c/a)

square root both sides

x + (b/(2a)) = sqrt(((b^2)/(4a^2)) - (c/a))
or
x + (b/(2a)) = sqrt(((b^2)/(4a^2)) - ((4ac)/(4a^2)))
x + (b/(2a)) = sqrt((b^2 - 4ac)/(4a^2))

x + (b/(2a)) = (sqrt(b^2 - 4ac))/(2a)

subtract (b/(2a)) from both sides

x = (-b/(2a)) ± ((sqrt(b^2 - 4ac))/(2a))
x = (-b ± sqrt(b^2 - 4ac))/(2a)

just in case you didn't know how it works.

(-b/(2a)) gives you the "x" value of the vertex

y = a(-b/(2a))^2 + b(-b/(2a)) + c
y = a(((-b)^2)/(4a^2)) + ((-(b^2)/(2a))) + c
y = ((-ab^2)/(4a^2)) + ((-(b^2))/(2a))) + c
y = (((-b)^2)/(4a)) + ((-(b^2)/(2a)) + c
y = (((-b)^2)/(4a)) + ((-2(b^2))/(4a)) + ((4ac)/(4a))
y = ((-b)^2 - 2b^2 + 4ac)/(4a)
gives you the "y" value of the vertex.

and "c" tells you the y-intercept

when following the completing the square method you can see where the -2a came in at.

So i can only tell you how to find the x-intercept(s), the y-intercept, and the vertex.