Question

vertex for the graph of y = 2x2 + 12x + 10

Answer 1

http://www.mathwarehouse.com/geometry/parabola/standard-and-vertex-form.php
CHECK THIS SITE OUT!! :D

Answer 2

write in the form( x-h)^2=4a(y-k)
then (h,k) is the vertex.

Answer 3

@surjithayer I don't recognize that form?? Can certainly use vertex form which is

\(y=a(x-h)^2 + k\) and then (h,k) is the vertex.

OR you can just use the fact that the x-coordinate of the vertex is given by:

\(x=\dfrac{-b}{2a}\)

to get the x-coordinate, substitute that back into the equation for the y-coordinate.

Answer 4

y=2(x^2+6*x+5)
implies y=2[(x+3)^2-4]
(y+8)/2=(x+3)^2;
a=1/8;
iumplies 4*1/8(y+8)=(x+3)^2
vertex is (-3,-8)

Answer 5

If you rearrange the vertex form, you can get

\(\dfrac{y-k}{a}=(x-h)^2 \)

although I've never seen it used that way. :)

Answer 6

2(x^2+6x+5)=y
2(x^2 +6x+(3)^2-(3)^2+5)=y
2((x)^2+6x+9-9+5)=y
2(x+3)^2-8=y
2(x+3)^2=y+8
(x+3)^2=1/2 (y+8)
vertex is (-3,-8)

Answer 7

4a=1/2
a=1/8

Answer 8

I'm still not following what your "a" is referring to. The standard form for a quadratic is

\(ax^2+bx+c=0\) and here, the equation has a=2.

Once you had: \(2(x+3)^2-8=y\) above, that is what is usually taught as "vertex form"
\[y=a(x-h)^2 + k\] where a is the coefficient on the leading term; and (h,k) is the vertex (as you found).

Using this as our understanding of "a", what you are calling a in the rearragement to \(4a(y-k)=(x-h)^2\) would be \(\dfrac{1}{4a}\).

Then:
\(4\cdot\dfrac{1}{4a}(y-k)=(x-h)^2\)
\(\dfrac{1}{a}(y-k)=(x-h)^2\)
\((y-k)=a(x-h)^2\)
\(y=a(x-h)^2+k\)

And when we solve:

\(\dfrac{1}{4a}=\dfrac{1}{8}\) you get a=2, which is as we expect, the leading coefficient.

Answer 9

These are all types of parabola with vertex (0,0) and focus at F