free medal

not really

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

Question

free medal




not really 


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

imstuck · Answer

It is a parabola with its center at (1,0).

jenniferjuice · Answer

ok so what do i do now?

imstuck · Answer

|dw:1406085328556:dw|

imstuck · Answer

The standard form of this parabola is $$y=4px ^{2}$$The value of the p is found by counting the number of units the focal point is from the center.  Here p = 1.  So our equation, taking into consideration that the center is at (1,0) is$$y=4(x-1)^{2}$$

imstuck · Answer

|dw:1406085498233:dw|

imstuck · Answer

That's it!

jenniferjuice · Answer

wait so its d?

imstuck · Answer

If you need it in standard form as opposed to graphing form you would have$$y=4x ^{2}-8x+4$$

imstuck · Answer

oh...my bad.  I figured p wrong.  4p=1 so p = 1/4.  sorry.

imstuck · Answer

yes, its the last one of your choices.  Sorry about the wrong value for p.  TY for showing it to me!

jenniferjuice · Answer

thank you for your help though :)

imstuck · Answer

TY for the medal...and those certainly are not free medals!  Those darn quadratic equations are stinkers!