Question

Identify the vertex and the y-intercept of the graph of the function (I'm posting the picture problem)

Answer 1

a. vertex: (3,2); y-intercept: -16
b. vertex: (3,-2); y-intercept: -18
c. vertex: (-3,-2); y-intercept: 11
d. vertex: (-3,2); y-intercept: -16

Answer 2

I'm used to solving these problems by writing the function in transformational form. Are you familiar with that?

Answer 3

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

Answer 4

Oh yea of course

Answer 5

to find the y-intercept, simply set x = 0, and solve for "y"

Answer 6

If you write the function in transformational form, you'll get something like this\[y-VT=\frac{ 1 }{ VS }\left( x-HT \right)^2\]VT (vertical translation is the y-coordinate of the vertex and HT (horizontal translation) is the x-coordinate of the vertex. As jdoe0001 syas, the y-intercept is where x=0, so substitue 0 in for x and solve for y.

Answer 7

ok so the equation will be y=-2(0+3)^2 +2?

Answer 8

When finding the y-intercept, yes.

Answer 9

ok give me a minute and i'll let you know what i got

Answer 10

i got 38...

Answer 11

@ospreytriple

Answer 12

Not what I get. Remember, it's only the quantity in the parentheses that's squared...the 2 in front doesn't get squared.