Question

@phi an add-on to this question is this..

Use the method of completing the square to write the equation of the given parabola in this form: (y-k) = a(x-h)^2 where a is not equal to 0.

Find the vertex of this parabola:
y = 5x^2 + 10x - 5

Answer 1

when you have time, see
http://www.khanacademy.org/math/algebra/quadratics/completing_the_square/v/solving-quadratic-equations-by-completing-the-square

in the meantime, this one is messy because of the 5 in front of the x^2
I would "factor out the 5) from each term. can you do that ?

Answer 2

Wouldn't you multiply 5 * 10? Or I'm lost. Lol

Answer 3

they want you to complete the square. If you don't know how to do that, the video explains how. But the first step is factor 5 from each term.

For example, if you have 2(x+1) you can distribute the 2 to get 2x + 2
you "undo" that by "factoring" 2 out of each term to get 2(x+2)

use that idea with 5x^2 + 10x - 5

Answer 4

divide each term by 5 and put the 5 on the outside of some parentheses
\[ 5x^2 + 10x - 5 = 5(x^2 +2x -1 ) \]

Answer 5

as a check, distribute the 5:
\[ 5(x^2 +2x -1 ) = 5x^2 +10x -5 \]
which is the original

Answer 6

completing the square is a procedure that takes some time to explain why it works.
but once you have the form
x^2 +bx+c
take the "b" , divide it by 2, and square it, to get \( \frac{b^2}{4} \)
now we use a trick. notice that \( \frac{b^2}{4} - \frac{b^2}{4} =0\)
we can add zero to anything and not change it. here we do this:
\[ x^2 +bx+ \frac{b^2}{4} - \frac{b^2}{4} + c \]
it turns out that
\[ x^2 +bx+ \frac{b^2}{4} = \left(x + \frac{b}{2} \right)^2\]

Answer 7

for your problem
\[ y=5(x^2 +2x -1 ) \]
b is "2" (in front of the x)
divide by 2 to get 2/2=1
square it: 1*1= 1
now add 1-1 (which is zero)
\[ y=5(x^2 +2x +1 - 1-1 ) \]
the \( x^2 +2x +1 = (x+1)^2 \) so you now have
\[ y=5( (x+1)^2 −1−1) \]
-1 -1 simplifies to -2:
\[ y=5( (x+1)^2 −2) \]

the last step is distribute the 5
\[ y= 5(x+1)^2 -10 \]