SOLUTION: I was wondering if I solved this problem right. I was supposed to write the equation in vertex form. y=2x^2+12x+13 y-13=2x^2+12x y-13=2(x^2+6x) y-13=2(x^2+6x+9-9) y-13=2(x+3

Algebra ->  Linear-equations -> SOLUTION: I was wondering if I solved this problem right. I was supposed to write the equation in vertex form. y=2x^2+12x+13 y-13=2x^2+12x y-13=2(x^2+6x) y-13=2(x^2+6x+9-9) y-13=2(x+3      Log On


   

Question 632946: I was wondering if I solved this problem right. I was supposed to write the equation in vertex form.
y=2x^2+12x+13
y-13=2x^2+12x
y-13=2(x^2+6x)
y-13=2(x^2+6x+9-9)
y-13=2(x+3)^2-9
y-13=2(x+3)^2-18
y=2(x+3)^2-5?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
looks like you got it right.
the procedure i know is like this:
y = 2x^2 + 12x + 13
set y = 0 to get:
2x^2 + 12x + 13 = 0
subtract 13 from both sides of the equation to get:
2x^2 + 12x = -13
factor out the 2 to get:
2 * (x^2 + 6x) = -13
complete the square of x^2 + 6x to get:
2 * ((x+3)^2) - 9) = -13
distribute the multiplication on the left side of the equation to get:
2 * (x+3)^2 - 18 = = - 13
add 13 to both sides of the equation to get:
2 * (x+3)^2 - 5 = 0
replace 0 with y to get:
y = 2 * (x+3)^2 - 5
you did essentially the same thing and got the same answer.
to confirm, i graphed both equations to see if they were identical.
they were.
the graph of both equations is shown below:
graph%28600%2C600%2C-10%2C10%2C-10%2C30%2C2x%5E2%2B12x%2B13%2C2%28x%2B3%29%5E2-5%29
in this graph i added 1 to the original equation to show you that they are both there.
the graph of both equations with 1 added to the original equation is shown below:
graph%28600%2C600%2C-10%2C10%2C-10%2C30%2C2x%5E2%2B12x%2B13%2B1%2C2%28x%2B3%29%5E2-5%29