Question

express the equation in vertex form by completing the square
y=2x^2+4x

Answer 1

Vertex at (-1,-2)

Answer 2

how'd you get that?

Answer 3

You would need to complete the square. This means that we first factor out the 2, leaving us with 2(x^2+2x). Then we take half the coefficient of the bx or 2x. On side of your paper expand (x+1)^2. This gives us x^2+2x+1. Now that we have that, we see that the inside the parantheses of 2(x^2+2x) is missing the term 1, thus we add 1 and then subtract 1, giving us 2(x^2+2x+1-1). The then recognize that inside the parentheses we have (x+1)^2, so we write: 2(x+1)^2-1. We distribute the 2 to the -1 and get: (x+1)^2-2.

Answer 4

hold the phone. you can certainly do this if you like

Answer 5

or you can use the magic formula for finding the vertex, namely the first coordinate is
\[-\frac{b}{2a}\] and the second coordinate is what you get when you replace x by that number

Answer 6

woa

Answer 7

in this case
\[a=2,b=4,-\frac{b}{2a}=-\frac{4}{4}=-1\]

Answer 8

that is the first coordinate. now replace x by -1 and get the second coordinate.
\[2(-1)^2+4\times -1=2-4=-2\] finish

Answer 9

thank-you

Answer 10

express the equation in vertex form by "completing the square" y=2x^2+4x

Answer 11

vertex form is
\[2(x+1)^2-2\]

Answer 12

emphasis on the completing squizzle part

Answer 13

you can certainly do that if you like. or you can say it is going to be
\[2(x-h)^2+k\] where
\[h=\frac{b}{2a}\] and k is what you get when you replace x by
\[-\frac{b}{2a}\] in the original equation. keeps you from having to keep track of what you add and subtract etc. it keeps track for you

Answer 14

in fact however i made a mistake. it should be
\[2(x+h)^2+k\] where
\[h=\frac{b}{2a}\] sorry

Answer 15

yeah