Question

What is the equation of the quadratic graph with a focus of (1, 1) and a directrix of y = −1?

f(x) = −one fourth (x − 1)2 + 1

f(x) = −one fourth (x − 1)2

f(x) = one fourth (x − 1)2 + 1

f(x) = one fourth (x − 1)2

Answer 1

If it has a focus and a directrix, it is a parabola.

Answer 2

On a graph it looks like this:

Answer 3

The vertex is directly between the focus and the directrix, so it is located here:

Answer 4

A parabola ALWAYS wraps itself around the focus, so it opens upwards. That makes it an x^2 parabola.

Answer 5

The general equation for this type of parabola is \[y=4p(x-h)^{2}+k\]where p is the number of units the focus is from the vertex, h represents the x coordinate of the vertex, and k represents the y coordinate of the vertex.

Answer 6

so our focus is at (1,1) and our vertex is at (1,0), so the focus is 1 unit from the vertex in the positive direction (this is also how we know that this is a positive parabola). So p = 1, and the x coordinate of the vertex is 1 and the y coordinate is 0, so filling in our equation, we get this, before simplification:

Answer 7

\[y=4(1)(x-1)^{2}+0\]or, simplified, \[y=4(x-1)^{2}\]

Answer 8

some books will have you divide by the 4 and get \[\frac{ 1 }{ 4 }y=(x-1)^{2}\]but it depends upon the book.

Answer 9

Sorry, mixed up my values...your answer is the last one.

Answer 10

ok perfect man I have like three of those questions now I have a better idea on how to do it thanks @IMStuck

Answer 11

forgot to solve for p when I did that step. To find p, you use the equation 4p = (the coefficient on the x, which is a 1). 4p = 1, p = 1/4

Answer 12

you're very welcome!

Answer 13

ok thanks

Answer 14

These are by far the very hardest thing I taught to my Algebra 2 class last year. They struggled like mad!!!

Answer 15

I feel them @IMStuck

Answer 16

TY for the medal! Keep working with them! Have you gotten to hyperbolas yet? Oh boy on those!

Answer 17

not yet