1. ## help with precal problems plz

1. complete the square to write the equation of the parabola f(x) = -8x^2 - 24x + 15 in (h,k) form.

I can up with:
-8(x-3/2)^2 -3
( 3/2 , -3) ... but this is wrong... I dunno why...

2. find the domain of f(g(x))
f(x) = 1/(x-2) g(x) = 1/(radical x-3)

I came up with:
1/{[1/(radical x-3)] -2 } as the equation and so i know that x ≥ -3 and i also set up the equation (radical x-3) ≠ 1/2 because 1/(1/2) is 2 and 2-2 is 0.
so... how would I write the domain in the (x:x) = (-infinity, a number)(a number, infinity) form?

3. find a relationship between x and y so that (x,y) is equidistant from two points (5,2) and (-1,7). solve for a simplified relationship between x and y, don't solve for x and y.

I had no idea how to solve this one.

2. Your -8(x-3/2)^2 = -8x^2 -8.2.(-3/2)x - 8(-3/2)^2 = -8x^2 + 24x - 18.
You wanted -8(x+3/2)^2 which is -8x^2 - 24x - 18.

3. Originally Posted by u317d

2. find the domain of f(g(x))
f(x) = 1/(x-2) g(x) = 1/(radical x-3)

I came up with:
1/{[1/(radical x-3)] -2 } as the equation and so i know that x ≥ -3 and i also set up the equation (radical x-3) ≠ 1/2 because 1/(1/2) is 2 and 2-2 is 0.
so... how would I write the domain in the (x:x) = (-infinity, a number)(a number, infinity) form?
f(g(x))=1/(g(x)-2) is defined for g(x) ≠ 2 i.e, for (radical x-3) ≠ 1/2 ie
x ≠ 3+1/4=13/4
and g(x) is defined only when x>3
over all now, x has to be a number from 3 to 13/4 or 13/4 to infinity (excluding 13/4)
which we can write as domain = {x: x belongs to (3, 13/4) U (13/4, infinity)

4. Originally Posted by u317d

1. complete the square to write the equation of the parabola f(x) = -8x^2 - 24x + 15 in (h,k) form.
What is (h,k) form?

RonL

5. (h, k) usually refers to the coordinates of the vertex and the form

(x - h)² = 4a(y - k)

y = -8x² - 24x + 15

8x² + 24x = 15 - y
x² + 3x = 1/8(15 - y)

x² + 3x + 9/4 = 1/8(33 - y)

(x + 3/2)² = 1/8(33 - y)

(h, k) = (-3/2, 33)