Question

When y = 2x^2 + 5x + 10 is converted to the form y = a(x-p)^2 + q, the value of q is...

Answer 1

http://www.mathwarehouse.com/geometry/parabola/images/standard-vertex-forms.png what do you think?

Answer 2

I understand how to get standard form into vertex form, but the numbers are throwing me off.

Answer 3

hmmm well, you don't have any numbers in this case, the question is just asking
what is "q" as a matter of vertex coordinate

not asking for an absolute number, just what does "q" stand for in the vertex form

Answer 4

hh... ohh.... shoot.... well it does require you to change it... ok...so you'd need the vertex form... so... why it's throwing you off?

Answer 5

Well, you have to divide 5x by 2 and then continue on with all of the steps, I tried it multiple times and never got the correct answer. It is asking for the value of "q", and it is a decimal number

Answer 6

right

Answer 7

In the end, the answer is 6.875 for the value of q, and I would like to know how to get there

Answer 8

right... gimme a sec

Answer 9

Alright, thanks for helping me out!

Answer 10

\(\bf y = 2x^2 + 5x + 10\\ \quad \\ \quad \\
\textit{let us complete the square first}\\
y = (2x^2 + 5x) + 10\qquad \textit{now let's take common factor}\\
y = 2\left(x^2 + \cfrac{5}{2}x+\square^2\right) + 10\\ \quad \\
2n = \cfrac{5}{2}\implies n = \cfrac{5}{4}\qquad thus\\ \quad \\
y = 2\left(x^2 + \cfrac{5}{2}x+\left(\cfrac{5}{4}\right)^2\right) + 10\\ \quad \\

\textit{notice, we really added}\quad 2\times \left(\cfrac{5}{4}\right)^2\quad \textit{thus we also have to subtract it}\\
\textit{since all we're really doing is borrowing from "zero"} 0\\
y = 2\left(x^2 + \cfrac{5}{2}x+\left(\cfrac{5}{4}\right)^2\right) + 10-2\times \left(\cfrac{5}{4}\right)^2
\)

Answer 11

Ohh, I'm starting to see it now, thank you so much for helping me!

Answer 12

\(\bf y = 2\left(x^2 + \cfrac{5}{2}x+\left(\cfrac{5}{4}\right)^2\right) + 10-2\times \left(\cfrac{5}{4}\right)^2\\ \quad \\
y = 2\left(x^2 + \cfrac{5}{2}x+\left(\cfrac{5}{4}\right)^2\right)+10-2\times\cfrac{25}{16}\\ \quad \\
y = 2\left(x^2 + \cfrac{5}{2}x+\left(\cfrac{5}{4}\right)^2\right)+10-\cfrac{25}{8}\\ \quad \\
y = 2\left(x+\cfrac{5}{4}\right)^2+\cfrac{80}{8}-\cfrac{25}{8}\implies y = 2\left(x+\cfrac{5}{4}\right)^2+\cfrac{55}{8}\)

Answer 13

yw

Answer 14

Thank you!