Converting Standard to Vertex Form (Parabolas)?

Hi i would like to know how you convert a Standard Parabola (y = ax^2 +bx + c) equation into a Vertex Parabola Form (y = a(x-h)^2 +k)

Converting Standard to Vertex Form (Parabolas)?
y = ax^2 + bx + c





Your first step would be to factor "a" out of the first two terms.





y = a(x^2 + (b/a)x) + c





What you want to do is change what is currently in the brackets into a square binomial. This is accomplished by taking "half squared" of the coefficient of x, and adding this value within the brackets. By "half squared", I mean taking 1/2 of b/a, and then squaring it.





Calculation of "half squared" of b/a goes as follows:


a) Multiply by 1/2. 1/2 * b/a = b/2a


b) Square the numerator and denominator.


(b/[2a])^2 = (b^2)/[4a^2]





Now we add this value within the brackets.





y = a(x^2 + (b/a)x + (b^2)/[4a^2]) + c + ?





Note that I put the question mark there. This is because we added a new value in there, and we have to offset it because we're not allowed to add something to an equation without some sort of offsetting. Since we added (b^2)/(4a^2), we have to offset it; but remember, the actual value we added was [a * (b^2)/(4a^2)], or [ (b^2)/4a ] , so we have to SUBTRACT this value.





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





Now, we can factor the now square binomial.





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





And now we can put the last two terms under a common denominator.





y = a(x + b/(2a))^2 + [4ac -b^2]/[4a]





And that's the vertex form; note that





h = -b/2a, and


k = [4ac -b^2]/[4a]
Reply:in standard form, the vertex has x value = -b/2a. From that you can find the y value. You now have (h,k) in the vertex form.