# Finding Solutions and Graphing F(X)

• March 4th 2010, 02:10 PM
Finding Solutions and Graphing F(X)
Greetings,

Hello again. I've got a few more questions this week. I've got another worksheet of problems and I need to graph them, however with these types of problems I'm not really sure how to convert the answers into something I can graph. I'm also not too sure about the f(x) thing, which I'm not entirely positive how to solve.

All of the problems below I need to graph...

1. $f(x) = x^2-6x+4$
I'm not entirely sure of myself on the f(x) concept. I'm probably wrong, but it seems like my instructor said that f(x) can be replaced with 0, but I'm probably wrong or maybe I just got part of what he was saying.

2. $g(x)=x^2+3x-10$
Is g(x) is the same as f(x), just using a different letter, or is it a different problem entirely?

3. $g(x)=-(x+2)^2$
If I understand it correctly, I'd just multiply everything in the parenthesis by -1 (-x, -2) and than have another set of the same numbers and multiple them together, correct? So (-x-2)(-x-2)... And than somehow formulate something to graph once I finish with the answer?

So, anyway, those are the ones I'm struggling with. Any help or advice would be much appreciated.
Thank you!
• March 4th 2010, 03:24 PM
Ben92
Hello!

You should convert the quadratic equation to an another form, like:

$(x-3)^{2}-5=x^{2}-6x+4$

You can see that the point is $x^{2}-6x$, so you can use the $\left(a-b\right)^{2}$ identity.

If you done with this tranformation you will have an $x^{2}$ function, shift it with 3 on the $x$ axis, and subtract 5 from each value on the $y$ axis.
• March 5th 2010, 04:02 AM
HallsofIvy
Quote:

Greetings,

Hello again. I've got a few more questions this week. I've got another worksheet of problems and I need to graph them, however with these types of problems I'm not really sure how to convert the answers into something I can graph. I'm also not too sure about the f(x) thing, which I'm not entirely positive how to solve.

All of the problems below I need to graph...

1. $f(x) = x^2-6x+4$
I'm not entirely sure of myself on the f(x) concept. I'm probably wrong, but it seems like my instructor said that f(x) can be replaced with 0, but I'm probably wrong or maybe I just got part of what he was saying.

f(x) can be replaced by 0 if the problem is to find where (what value of x) the function value is 0. The better way to graph such a function is to replace x with several different values, find the corresponding values of f(x), graph those (x, f(x)) points and the "connect the dots".

Quote:

2. $g(x)=x^2+3x-10$
Is g(x) is the same as f(x), just using a different letter, or is it a different problem entirely?
Of course, you can call a function anything you like but what do you mean by "the same as f(x)". It is certainly NOT the same as the previous problem because the coefficients are different.

Quote:

3. $g(x)=-(x+2)^2$
If I understand it correctly, I'd just multiply everything in the parenthesis by -1 (-x, -2) and than have another set of the same numbers and multiple them together, correct? So (-x-2)(-x-2)... And than somehow formulate something to graph once I finish with the answer?
I have no idea what you mean by this! What do you mean by "multiply by "-1(-x, -2)"? Since that is neither a number nor an algebraic expression, how would you multiply?

To graph it, choose several different values of x, for each value, add 2, then square the result, and finally multiply by -1. That gives you the value of f(x) for that x- mark the points (x, f(x)) and "connect the points".

Quote:

So, anyway, those are the ones I'm struggling with. Any help or advice would be much appreciated.
Thank you!
• March 5th 2010, 04:01 PM
Alright, so I'm just trying to make sure I understand the f(x) thing, with graphing it, too.

So for a problem with f(x) I'd just substitute f(x) with three different numbers and solve the problem? Here's an example, so I can see if I'm understanding it...

$f(x)=x^2-6x+4$
$X=1, 2, 3$

$1=1^2-6(1)+4$
$1=1-6+4$
$1=-1$
$X=1, Y=-1$

And than I'd just replace 1 with 2, than 3, and so on. Is that correct? Or do I just change f(x) to a number and leave the rest of the problem the same? Something like...
$1=x^2-6x+4$
$0=x^2-6x+4-1$
But this way would give me an answer that I have no clue how to translate into a point I could graph...
$\frac{6+-\sqrt{48}}{2}$

Did I do anything here right?
• March 5th 2010, 04:06 PM
Alegend
If there's something that I learned from Algebra I For Dummies, it's how to graph a quadratic equation. You set the equation equal to y, and graph.
• March 6th 2010, 03:47 AM
HallsofIvy
Quote:

If there's something that I learned from Algebra I For Dummies, it's how to graph a quadratic equation. You set the equation equal to y, and graph.
I'm sorry but what are you saying here? You learned to graph a function by graphing it?