Math Homework Question (algebraic functions)

meatflower

This is actually a take home test that's due in 8 hours and I'm scrambling to finish it.

I'm sure the solution to this is really simple but I'm not finding anything on it in my book or in my notes. Any help would be appreciated.

Given f(x)= -x^2 - x - 5

Write f(x) in the form f(x)= a(x-h)^2 + k

I have never seen that form before and have no idea what h and k are. I'm assuming a is -1, but that still leaves me with the problem of what h and k are.

If anyone could tell me what that form is called that would also help.

Thank you.

piL

For this, you must complete the square, which is pretty much using part of the formula to make something you can factor.

f(x)= -x^2 - x - 5
f(x)= -1(x^2 + x + 5)
f(x)= -1((x^2 + x + 1/4) + 19/4)
f(x)= -1((x+1/2)^2 + 19/4)
f(x)= -(x+1/2)^2 - 19/4

a = -1
h = -1/2
k = -19/4

(-1)*(x-(-1/2))^2 + (-19/4)

Describing the concept of completing the square by working backwards:

a(x-h)^2 +k

a(x^2-2hx+h^2) + k

ax^2 - 2ahx + ah^2 + k

Then, if you take the normal formula to be

mx^2 + nx + c

Then

mx^2 + nx + c = ax^2 - 2ahx + ah^2 + k

and you can conclude

m = a

n = 2ah

c = ah^2 + k



The process of completing the square basically is to take whatever number you need to create a factorable number from the constant that's included in the equation.

Hope that helps. This might help a bit.

meatflower

Thank you!

I know how to complete the square...my brain is just so fried I'm not thinkin straight. Waiting to do all the homework+the take home test the night before it's due was not the brightest of ideas I've had lately.

Thanks again.

piL

No problem! Glad I could help.

I've been doing math for a while now, (read: years--no need to spout credentials), and I've only just now started to remember to use the complete the square technique. I used to always overlook it, time and time again. I probably learned about it 7 years ago.

It's wierd, but I used to always forget about it. Now I just always forget about other things.

meatflower

I have one more question, if you don't mind.

I'm being asked to find the inverse of the following function:

f(x)= (4 - 2x)/(3 - 6x)

and have come up with:

f^-1(x)= (3x - 4)/(-2 + 6x)

I've checked it twice and it seems right but then when I check f(x) composed with it's inverse I don't get x, I get 4/3 - 1/3x + 1/3. That means it's not really the inverse function.

The only explanation I coudl come up with was that the original function isn't one to one or my calculation of the inverse function is wrong. Since f(x) looks like it's one to one to me I'm guessing it's an error on my part.

Thanks again.

piL

You were close! To solve for the inverse, you substitute the f^-1(x) into the equation. That's a bitch to write, f^-1(x) = g for purposes of this forum typing exercize. Plus, I like g anyway, because you can write it without having to write extra shit like the ^-1 and (x) crap while you're trying to solve an equation that may or may not have real exponents and values of x in there

First, replace all x values with the substituted value (g), and then set this f(g) = x.

Then solve for g.

f(g) = (4-2g)/(3-6g) = x

Multiply both sides by (3-6g):
(4-2g) = x(3-6g)

Distribute:
4-2g = 3x - 6gx

Subtract 3x from both sides and add 2g to both sides to get all g terms on one side:
4-3x = 2g - 6gx

Condense:
4-3x = g(2-6x)

Divide both sides by 2-6x to solve for g:

(4-3x)/(2-6x) = g = f^-1(x)

meatflower

Awesome!

Apparently starting the equation like this:

g = 4 - 2x / 3 - 6x

as opposed to

4 - 2x / 3-6x = g

makes all the difference in the world.

I really should know better than this.

You've really saved me on this whole thing, thank you once again.

piL

meatflower wrote:

Apparently starting the equation like this:

g = 4 - 2x / 3 - 6x

as opposed to

4 - 2x / 3-6x = g

makes all the difference in the world.

I think you meant:

g = 4-2x/3-6x

as opposed to
4-2g/3-6g = x


Or atleast I hope you did

Anyway, plenty glad enough to help.