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humorist-workshop
mathhhhh :(
I suck at math and I need a lot of help. I have some homework and I've done part of it but I dont understand these problems:
Each of the next questions refer to the parabola y = 5(x+2)^2 + 7.
Find the vertex, focus, and length of the latus rectum of the parabola
The directrix line of the parabola is the line y =
Find the standard form of the ellipse given by the equation x^2 + 25y^2 -2x + 150y + 201 = 0
Here is the equation of an ellipse in standard form: (1/16)(x + 2)^2 + (1/9)(y - 5)^2 = 1 . Each of the next questions refer to this ellilpse.
Find the center, foci, length of the major axis, and the distance between the two foci of the ellipse
Find the standard form of the hyperbola given by the equation 4x^2 - 25y^2 - 50y - 125= 0
Here is the equation of a hyperbola in standard form: (1/10)(x - 1)^2 - (y - 1)^2 = 1 . Each of the next questions refer to this hyperbola.
Find the center, the two asymptotes, and the foci of the hyperbola
One of the two axes (transverse or conjugate) is parallel to the y-axis. Find its length
Find the distance between the two points (3,-4) and (-4,5)
Even if you cant help answer all of these a few would really help, Im super confused.. and maybe someone on here can help because Im sooo bad at algebra.
Thanks:)
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Want to answer more questions in the Work & School category?
Maybe give some free advice about: School?
I'm not gonna give you answers (I'm not here to do homework for you), but I'll try to help. Honestly, you should try looking in your textbook. We're not going to explain it any better than your textbook is.
I don't remember too much about this part of math (it's been a while), so I can't help on every problem.
>> y = 5(x+2)^2 + 7
Finding the vertex is extremely simple.
If you know calculus (You probably don't, but):
Find the derivative, set it equal to 0, and solve for x (This is your x-coordinate). Then plug the x value into the original equation and solve for y to obtain the y-coordinate.
If you dont' know calculus:
1) put the equation into y = ax^2 + bx + c form.
2) plug your a, b, and c values into this equation:
x = -b / 2a
Whatever you get for x is your x-coordinate.
3) plug your newly-found x value into the original quadratic equation and solve for y. That's your y-coordinate of the vertex.
>> "Find the standard form of the ellipse given by the equation x^2 + 25y^2 -2x + 150y + 201 = 0"
You should have been able to figure this out simply by looking at the next question about the ellipse in standard form.
The standard form of an ellipse is:
(x-h)^2/a^2 + (y-k)2/b^2 = 1
The center of an ellipse can be determined by looking at the equation in standard form:
center = (h, k)
(h, a, k, and b are all constants. keep this in mind)
How to get from x^2 + 25y^2 -2x + 150y + 201 = 0 to standard form:
1) organize the equation. put the x terms together and the y terms together. put any constants on the other side of the equation. Use parentheses if it helps:
(x^2 -2x) + (25y^2 +150y) = -201
2) Complete the square for each group in parentheses. (If you dont' know how to complete the square: [Link](Mouse over link to see full location))
Your result should be in this form:
a(x-h)^2 + b(y-k)^2 = C [all letters here other than x and y are just constants. don't freak out. you will get a constant in front of the squared terms only if you factored a constant out after completing the square. make sure you DO factor if possible.]
3)divide by a and b.
4)If c does not = 1, divide by c.
>>"Find the distance between the two points (3,-4) and (-4,5)"
Distance formula!
D = square root of [ (x - x)^2 + (y-y)^2 ]
Note that by (x-x), I don't mean zero. It's the difference of the first x value and the second x value. Same goes for y.
[Link](Mouse over link to see full location)
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