# Math Help - Need help with functions!!

1. ## Need help with functions!!

For the function f(x) = x^2 - 2x + 1

(a) find f(0)

(b) solve f(x)=0

I don't get it!! Please explain!!

How would I graph y = f(x) = x^2 and x - y = 6????

I'm so confused!!!!

2. Originally Posted by poeticprincess
For the function f(x) = x^2 - 2x + 1

(a) find f(0)

(b) solve f(x)=0

I don't get it!! Please explain!!

How would I graph y = f(x) = x^2 and x - y = 6????

I'm so confused!!!!
Notice that in the notation "f(0)" 0 is replacing the "x" in f(x), this means that 0 is replacing every "x" in the function f(x) = x^2 - 2x + 1

In other words, since f(x) = x^2 - 2x + 1
f(0) = (0)^2 - 2(0) + 1 = 0 - 0 + 1 = 1

For the second part of the problem, notice that we are letting f(x) = 0. This means that instead of having:
f(x) = x^2 - 2x + 1
We are letting:
0 = x^2 - 2x + 1 ... (the 0 replaced the "f(x)")

To solve for x, we need to factor this quadratic and use the zero product property:
0 = (x - 1)(x - 1)
x - 1 = 0 --> x = 1 ... (we have (x - 1) twice, so we only need to set it equal to 0 once).

Therefore, we have that when f(x) = 0, x = 1.

3. Thnx a lot!

4. Originally Posted by poeticprincess
...

How would I graph y = f(x) = x^2 and[I][FONT=Palatino Linotype][FONT=Verdana] x - y = 6
...
Hi,

you have 2 functions. The easiest way to draw the graphs of function is to make a table of x-values with the corresponding y-values.

y = x²
Code:
  x         y
--------------
±3    |    9
±2    |    4
±1    |    1
0    |    0
--------------
The graph of the second function is a straight line:
x - y = 6 ==> y = x - 6
To draw this graph you only need 2 points:
Code:
 x       y
-----------
0   |  -6
6   |   0
-----------
As you may have noticed I have chosen the intercepts of the graph and the coordinate axes.

I've attached a diagram of both graphs.

5. Originally Posted by earboth
Quote:
Originally Posted by poeticprincess
[i]...

How would I graph y = f(x) = x^2 and[font=Palatino Linotype][font=Verdana] x - y = 6
...
Hi,

you have 2 functions. The easiest way to draw the graphs of function is to make a table of x-values with the corresponding y-values.

y = x²
Code:
  x         y
--------------
±3    |    9
±2    |    4
±1    |    1
0    |    0
--------------
I still don't really understand how you got the answer to this problem. If it's not too much trouble can you please explain this to me a little bit further. Thank you for any help you can give me.

6. Originally Posted by poeticprincess
I still don't really understand how you got the answer to this problem. If it's not too much trouble can you please explain this to me a little bit further. Thank you for any help you can give me.
Hi,

it doesn't make much sense if I start an essay about functions and their graphs and how to draw those graphs. First I have to know what you actually know about these subjects:

1. Do you know how to place a point in the coordinate plane if you know the coordinates of the point?

2. Do you know how to decribe a function by an equation?

7. Originally Posted by earboth
1. Do you know how to place a point in the coordinate plane if you know the coordinates of the point?

2. Do you know how to decribe a function by an equation?
Well I do know how to place points in the plane if I have the coordinates but I alwaysa get confused in describing a function by an equation. If you could please explain this to me I'd be very greatful. Thanks in advance!

8. Hello,

I didn't forget you. It only took some time to write the following text.
(I translated this text into English, so maybe you'll come across some expressions which are unusual to you - then ask again):

If two terms are connected by an equal sign you have an equation. If this equation contains one variable (usually x or t or ...) you can try to find a value for this variable such that the equation is a true statement. This specific value is called a solution of the equation.
Example:
15 – (1/3)*x = (4/5)*x – 2
You can try to find a solution by trial and error or you can transform this equation without changing the possible solution. I assume that you know methods to solve this equation for x. You should get:
x = 15.
If you plug in this value instead of x the equation becomes a true statement:
15 – (1/3)*15 = (4/5)*15 – 2
15 – 5 = 12 – 2
10 = 10 is obviously true.

If you've got an equation with two variables (usually x,y or r,t or ...) this method doesn't work. You'll get a set of pairs of numbers which satisfy the equation.
Example:
3x – 2y = 8
Now you can choose any value for x and calculate afterwards the corresponding value for y such that the equation becomes a true statement.
x = 0 then y = -4. You get the pair (0, -4)
x = 2 then y = -1. You get the pair (2, -1)
x = 10 then y = 11. You get the pair (10, 11)
etc. ...
You can solve the equation for y first:
y = (3/2)*x – 4
and then choose a value for x: x = 2
then y = (3/2)*2 – 4 = 3 – 4 = -1 and you get the pair (2, -1) as before.

As you can see the y-value depends on the x-value. Every x-value is assigned uniquely to one y-value. This unique assignment is called a function. (Personal remark: (1) I'm not certain if the expression "assignment" - meaning mapping – is used correctly here. (2) My explanations are not complete and they cover only a part of the basics about functions)

As you have seen, you get a set of pairs of numbers. Each pair of numbers represents a point in the coordinate plane. All pairs (x, y) which satisfy the equation of the function form the graph of the function. If you want to draw the graph of a function you have to calculate first some points (= you have to calculate the coordinates of some points), draw these points and afterwards try to get the shape of the complete graph by connecting the points adequately.

Now compare my first post: You'll see a table of x-values in the first column and the corresponding y-values in the second column. I've plugged in the x-values into the equation of the given function and calculated the y-value. Then I draw the points and connected them. That's all.

9. Thanks earboth! Now I actually understand what I'm doing. Thank you!