Question

What is an equation of a parabola with the given vertex and focus?

vertex: (-2, 5) ; focus (-2,6)

Answer 1

@nerds

Answer 2

lol

Answer 3

@jhonyy9 @zepdrix @agent0smith

Answer 4

Are you familiar with vertex form?

Answer 5

@legomyego180

Answer 6

yeah
i am

Answer 7

Can you start plugging some things in?

Answer 8

just cuz ur Human Calculator doesn't mean a common bird doesn't know vertex form



1 sec gotta google soemething :>

Answer 9

vertex form
The vertex form of a quadratic is given by
y = a(x – h)2 + k, where (h, k) is the vertex.
The "a" in the vertex form is the same "a" as
in y = ax2 + bx + c (that is, both a's have exactly the same value). The sign on "a" tells you whether the quadratic opens up or opens down. Think of it this way: A positive "a" draws a smiley, and a negative "a" draws a frowny. (Yes, it's a silly picture to have in your head, but it makes is very easy to remember how the leading coefficient works.)

In the vertex form of the quadratic, the fact that (h, k) is the vertex makes sense if you think about it for a minute, and it's because the quantity "x – h" is squared, so its value is always zero or greater; being squared, it can never be negative.

Suppose that "a" is positive, so a(x – h)2 is zero or positive and, whatever x-value you choose, you're always taking k and adding a(x – h)2 to it. That is, the smallest value y can be is just k; otherwise y will equal k plus something positive. When does y equal only k? When x – h, the squared part, is zero; in other words, when x = h. So the lowest value that y can have, y = k, will only happen if x = h. And the lowest point on a positive quadratic is of course the vertex.

If, on the other hand, you suppose that "a" is negative, the exact same reasoning holds, except that you're always taking k and subtracting the squared part from it, so the highest value y can achieve is y = k at x = h. And the highest point on a negative quadratic is of course the vertex.

Answer 10

i know what vertex form is

Answer 11

im smrt

Answer 12

Cool, I can tell.

\[y=a(x-h)^2+k\]

Where h is the x-value of your vertex, and k is the y value.

Answer 13

That's the vertex form, all right. But what are you going to do with the info regarding the location of the FOCUS? You'll need a different equation for that case.

Look up "equations of parabolas." Find the one that involves the constant "p."

Figure out the value of p from the original problem statement.

Show your work, please.