Results 1 to 5 of 5

Thread: Need help figuring out this linear equations table

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    10

    Need help figuring out this linear equations table

    X. F(x)
    4. 13
    2. 7
    0. 1
    -1. -2

    using the table, what is the value of f(-1)?

    Write a linear function equation for this table of values.

    Using this equation or the table itself, find the value of f(1)

    I don't understand this.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Mar 2009
    From
    Alberta
    Posts
    173
    Quote Originally Posted by brianfisher1208 View Post
    X. F(x)
    4. 13
    2. 7
    0. 1
    -1. -2

    using the table, what is the value of f(-1)?

    Write a linear function equation for this table of values.

    Using this equation or the table itself, find the value of f(1)

    I don't understand this.
    K so the question is saying that there is a function $\displaystyle y=f(x)$ such that plugging in any x into the function will give you a value for y=f(x) as listed in the table.

    Using this, we can think of the table as telling us this:

    $\displaystyle f(4)=13 $
    $\displaystyle f(2)=7$
    $\displaystyle f(0) = 1 $
    $\displaystyle f(-1) = -2$

    Now as for making a function. What can we decide from the table? It looks like if x =0, f(x)=f(0)=1. So it looks like we need to have a constant that isn't multiplied by any x.

    Next, lets looks at the values, how can we turn 2 into 7 and 4 into 13?

    Looks like $\displaystyle f(x) = 3x + 1$. Does this make sense how we came to this? It can be a bit of a trial and error process.

    Let's test it.

    $\displaystyle f(2) = 3(2) + 1 = 6 + 1 = 7 \ \ \ \text{Check!}$

    $\displaystyle f(4) = 3(4) + 1 = 12 +1 = 13 \ \ \ \text {Check!}$

    $\displaystyle f(0) = 3(0) + 1 = 0 +1 = 1 \ \ \ \text {Check!}$

    $\displaystyle f(-1) = 3(-1) + 1 = -3 +1 = -2 \ \ \ \text {Check!}$



    Can you find $\displaystyle f(1)$ now?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Bacterius's Avatar
    Joined
    Nov 2009
    From
    Wellington
    Posts
    927
    Nice avatar, Kasper !

    Otherwise, you might want to consider a more straightforward algebra´c way.
    Your problem strongly suggests that a linear relationship exists between $\displaystyle x$ and $\displaystyle f(x)$. This is equivalent to saying that $\displaystyle f(x) = ax + b$, for some $\displaystyle a$ and $\displaystyle b$, $\displaystyle a \neq 0$.

    1. Find $\displaystyle b$

    You have some values of the function (actually, only two are required). You can say that :

    $\displaystyle 13 = 4a + b$

    $\displaystyle 7 = 2a + b$

    We can slightly reformulate this :

    $\displaystyle 13 - b = 4a$

    $\displaystyle 7 - b = 2a$

    And now, we get smart ! We divide both equations together :

    $\displaystyle \frac{13 - b}{7 - b} = \frac{4a}{2a}$

    Which is equivalent to :

    $\displaystyle \frac{13 - b}{7 - b} = 2$

    Now we simplify the fraction :

    $\displaystyle 13 - b = 2(7 - b)$

    $\displaystyle 13 - b = 14 - 2b$

    $\displaystyle - b + 2b = 14 - 13$

    $\displaystyle \boxed{b = 1}$

    2. Find $\displaystyle a$

    Now that we know $\displaystyle b$, this is easy : we know that $\displaystyle b = 1$ : we can use a pair of values $\displaystyle \left ( x, f(x) \right )$ :

    $\displaystyle 13 = 4a + b$

    Since $\displaystyle b = 1$, we get :

    $\displaystyle 13 = 4a + 1$

    $\displaystyle 12 = 4a$

    $\displaystyle 3 = a$

    $\displaystyle \boxed{a = 3}$

    3. Conclusion

    You are done : you have found the linear relationship between $\displaystyle x$ and $\displaystyle f(x)$ :

    $\displaystyle \boxed{f(x) = 3x + 1}$

    Let's check those results :

    $\displaystyle f(4) = 4 \times 3 + 1 = 12 + 1 = 13$ -->

    $\displaystyle f(2) = 2 \times 3 + 1 = 6 + 1 = 7$ -->

    $\displaystyle f(0) = 0 \times 3 + 1 = 0 + 1 = 1$ -->

    $\displaystyle f(-1) = (-1) \times 3 + 1 = -3 + 1 = -2$ -->

    Now that we can find $\displaystyle f(x)$ from $\displaystyle x$ with a simple linear formula, let's answer the last part of the question : substitute $\displaystyle x = 1$ to find the value of $\displaystyle f(1)$ :

    $\displaystyle f(1) = 1 \times 3 + 1 = 3 + 1 = 4$

    Done !
    _________________

    Does it make sense ?
    Last edited by Bacterius; Mar 1st 2010 at 08:13 PM. Reason: Nice boxes
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, brianfisher1208!

    . . . $\displaystyle \begin{array}{c|c}x & f(x) \\ \hline\text{-}1 & \text{-}2 \\ 0 & 1 \\ 2 & 7 \\ 4 & 13 \end{array}$


    Using the table, what is the value of $\displaystyle f(-1)$ ?

    Um . . . -2 ?




    Write a linear function equation for this table of values.
    A linear function has the form: .$\displaystyle f(x) \;=\;ax+b$
    And we must determine $\displaystyle a$ and $\displaystyle b.$

    We can use any two values from our table.

    For example:

    . . $\displaystyle \begin{array}{ccccccccccccc}f(2) = 7: && a(2) + b &=& 7 && \Rightarrow && 2a + b &=& 7 & [1] \\
    f(4) = 13: && a(4) + b &=& 13 && \Rightarrow && 4a + b &=& 13 & [2] \end{array}$


    Subtract [2] - [1]: .$\displaystyle 2a \:=\:6 \quad\Rightarrow\quad a \:=\:3$

    Substitute into [1]: .$\displaystyle 2(3) + b \:=\:7 \quad\Rightarrow\quad b \:=\:1$


    Therefore: .$\displaystyle \boxed{f(x)\;=\;3x+1}$




    Using this equation or the table itself, find the value of $\displaystyle f(1).$

    $\displaystyle f(1) \;=\;3(1) + 1 \;=\;4$

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Bacterius's Avatar
    Joined
    Nov 2009
    From
    Wellington
    Posts
    927
    Interesting! As you can see, Brian, Soroban chose to solve by substitution a system of linear equations in order to find $\displaystyle a$ and $\displaystyle b$, while I decided to arrange the equations to allow a division that will get rid of an unknown, making solving easy. Two equivalent solutions to one problem (although mine was a bit longer).

    Thanks Soroban, I didn't know the command \boxed, I was sort of deceived my function didn't appear in math font
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Problems creating linear table
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Dec 14th 2011, 07:17 AM
  2. Problems creating linear table
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Dec 13th 2011, 07:10 PM
  3. using linear equation or pattern to complete table
    Posted in the Pre-Calculus Forum
    Replies: 0
    Last Post: Sep 20th 2009, 07:15 PM
  4. Replies: 1
    Last Post: Apr 5th 2009, 02:46 PM
  5. Replies: 0
    Last Post: Apr 2nd 2009, 12:07 AM

Search Tags


/mathhelpforum @mathhelpforum