Algebra 2 Problem help?

Question

How do you find the vertex, foci, line of symmetry, and intercepts of the parabola x = -5y^2 + 30y +11.
I understand how you do all that with equations with the Y=, but not with the X=. I really need help, I've been stuck with these kinds of X= problems for months. Thanks. And also, does anyone know how to find The equations for the asymptotes for a hyperbola? (separate problem)

Answer 1

If you just treat the symbol as just symbols without their meaning on a Cartesian co-ordinate system, then you can blithely go about doing these X= ______ the exact same way as you to Y= ____.
The graph of these functions x=f(y) also look like parabolas, but rotated clockwise by 90degrees. Now to find the y intercepts, let x=0 and so we have -5y^2+30y+11=0 - here use the quadratic eqution to solve for y.
y= [-30 +/- sqrt(30^2 - 4*(-5)(11))]/2(-5)

For the x intercept, let y=0, therefore x=11 is the x intercept.

The line of symmetry will be a horizontal line. You can find it by finding the halfway point between the two y intercepts. For example if for some parabola the y intercepts were -1 and 1 the midway point would be 0 and so the symmetry line is y=0.

To find the vertex the first method that comes to mind is to differentiate x= f(y) with respect to y, and then equate the derivative to 0. Then solve for y. This will give you the y co-ord of the vertex. Then sub back into the original parabola function to find the x co-ord.

I've never worked with foci of a parabola...not sure about that. Hope this helps!

Answer 2

Dont get confused with the 2 types of parabola

if u cn solve it by replacing the x and y, u shud be able to do this too .. its exactly the same method .... even the answers will be the same except that x and y will be exchanged .. try it that way.

as for asymptotes, suppose x^2/a^2 - y^2/b^2 = 1 is the hyperbola
then the asyptotes are
x/a - y/b = 0
x/a + y/b = 0

Answer 3

It's not all that difficult. Just turn the paper 90 degrees counterclockwise and you have a y=f(x) problem. I'm surprised that your text doesn't have the hyperbola asymptotes. Look again.

Answer 4

the area between 2 factors relies on the pythagorean theorem. You calculate the full upward thrust (distinction between the y coordinates) and the run (distinction between the x coordinates), and then build a top triangle such that the hypotenuse is the line between the two factors. -14 - (-2) = -12 (your "upward thrust") 12 - 9 = 3 (your "run") via pyth thm, (-12)^2 + 3^2 = d^2, the place d is the area between the two factors. (-12)^2 + 3^2 = d^2 one hundred forty four + 9 = d^2 153 = d^2 d = sqrt(153), or the sq. root of 153 = approximately 12.369. wish it helps.