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College Algebra Questions -
August 21st 2009, 07:03 PM

I just got done with my first week of college, and I'm in a College Algebra math class. To finish off the week, the teacher gave us some work to do over the weekend, and I have a bunch of questions.

Right now we're doing functions. So if this is something that you know well and can explain in a simplified manner, then that'd be a great help.

The first problem that I'm having trouble with is this:

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

a. 2
b. -2
c. -x

The objective of the problems are to "evaluate each function at the given values of the independent variable and simplify".

For f(2) and f(-2), I came up with 15/4. For (-x), I came up with 3x-1...not confident in that answer at all.

My second problem is in a graph, which I made here:

The x-intercepts are 1 and -3 and the y-intercept is -3. The bottom of the function starts at -4, just clarifying if that graph is too sloppy to understand.

What I need to figure out is:
a. the function's domain
b. the function's range
c. the x-intercepts
d. the y-intercepts
e. and the missing function values that are indicated by question marks: f(-2) = ? and f(2) = ?

For the domain, I put {x|-3≤x≤1}, and for the range I put {y|y≥-4}, but I'm not confident in those answers.

I don't understand e period, so how would I find the missing function values?

The last question that I have for now is another graph that has me completely stumped. It looks like this:

Again, I have to find the domain, range, x-intercept and y-intercept, and the missing function values that are indicated with a question mark from f(-1) = ? and f(3) = ?

Re: College Algebra Questions -
August 21st 2009, 07:20 PM

I'll take a stab at it...
f(x)=(4(2)²+1)/(2³)
=(4*4+1)/8 ---> 2*2*2=8
=(16+1)/8
=17/8

f(x)= (4(-2)²+1)/-2³
= (4*4+1)/-8
=-17/8

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

On the first picture:
I'm sorry. I just read what the roots were (x-intercepts)
from this we get the equation (x-1)(x+3)=x²+2x-3 which is the parabola's equation. The domain (x-values) is...x has to be greater than or equal to -3 and and less than or equal to 1.

x>-3 and x<1
1>x>3 is the domain
{x|1>x>3}

The range is y>3 {y|y>3}

This second graph looks like an absolute value graph. The parent function for an absolute value is y=|x|. Say it's x=-1. y=|1| and y=|-1|. This results in a mirror image across the y-axis. So if you fold the graph it's identical on both sides.

The equation of this graph is roughly
y=|x|+1 (as it's shifted up one from the parent function with a vertex at 0,0) To find out the x-intercept look at the graph. Where does it intercept the x-axis? Nowhere. So there is no x-intercept.

The y-intercept: Where does it intersect on the y-axis? That would be at (0,1) as you can see there. =)

The domain is just where the x-values begin and where they end (in layman's terms). So, the x values begin at x>0 and continue on for infinity. So, x>0 is the domain. And in set notation: {x|x>0}

The range is just where the y-values being and where they end. Well, the only point where there is even a valid y point is at y=1. So the range is y> 1 (as you can see, the values continue on from 1 to foreverrr thanks to that arrow) In set notation: {y|y>1}

For the missing values you can refer to the picture OR plug and chug. I'll plug and chug. =)

f(x)=|x|+1
=|-1|+1
=1+1
=2

f(x)=|3|+1
=|3|+1
=3+1
=4

I hope I could help and I apologize for any mistakes as I'm a little math rusty from the summer. PM me if you need any more help. =)

Live and let live.

Last edited by <:3 )~; August 21st 2009 at 07:50 PM.

Re: College Algebra Questions -
August 21st 2009, 08:48 PM

Because if you slide over -3 on the x-axis and see what lies above it in y-axis terms, that would be your y value.
Well, x=-1, x=3. That's the equation for the function so just plug them in and solve for y. =)

Re: College Algebra Questions -
August 21st 2009, 10:10 PM

Quote:

Because if you slide over -3 on the x-axis and see what lies above it in y-axis terms, that would be your y value.
Well, x=-1, x=3. That's the equation for the function so just plug them in and solve for y. =)

Sorry, I still don't quite understand.

Quote:

f(-2) = ? and f(2) = ?

For e, I have to find the missing values of the function f(-2) and f(2). Don't the -2 and 2 mean something because each graph problem, they have different numbers in the f(number).

Re: College Algebra Questions -
August 22nd 2009, 01:45 AM

Oh okay. To find -2 and 2 plug and chug. Plug in the number and solve!
f(x)=|x|+1
x=-2
f(x)=|-2|+1
=2+1
=3
f(x)=|x|+1
x=2
f(x)=|2|+1
f(x)=3
Basically it's showing how the absolute value symbol creates a graph that is symmetric on the y-axis.

Also I would like to add this information: f(x) mean the function of x. y is a function of x. Which means it changes as x changes basically. f(x)=y

Okay, on the thing you don't understand - remember plotting points? Slide over -3 on the x-axis and plot the first point you see on the function itself with a y coordinate by scaling your pencil upward vertically. Like that. =)

Live and let live.

Last edited by <:3 )~; August 22nd 2009 at 01:56 AM.

Re: College Algebra Questions -
August 22nd 2009, 03:32 AM

Quote:

Originally Posted by <:3 )~

Oh okay. To find -2 and 2 plug and chug. Plug in the number and solve!
f(x)=|x|+1
x=-2
f(x)=|-2|+1
=2+1
=3
f(x)=|x|+1
x=2
f(x)=|2|+1
f(x)=3
Basically it's showing how the absolute value symbol creates a graph that is symmetric on the y-axis.

Also I would like to add this information: f(x) mean the function of x. y is a function of x. Which means it changes as x changes basically. f(x)=y

Okay, on the thing you don't understand - remember plotting points? Slide over -3 on the x-axis and plot the first point you see on the function itself with a y coordinate by scaling your pencil upward vertically. Like that. =)

I figured it out, but in a different way. I'm good from here on. Thanks for your help though.

Re: College Algebra Questions -
August 22nd 2009, 06:09 PM

No problem! I'm glad that you figured it out. It's kinda difficult to explain it in text and not be showing somebody it. Best of luck in that algebra class x