# Math Help - Need help figuring out this linear equations table

1. ## 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.

2. Originally Posted by brianfisher1208
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 $y=f(x)$ 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:

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

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

Let's test it.

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

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

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

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

Can you find $f(1)$ now?

3. Nice avatar, Kasper !

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

1. Find $b$

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

$13 = 4a + b$

$7 = 2a + b$

We can slightly reformulate this :

$13 - b = 4a$

$7 - b = 2a$

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

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

Which is equivalent to :

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

Now we simplify the fraction :

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

$13 - b = 14 - 2b$

$- b + 2b = 14 - 13$

$\boxed{b = 1}$

2. Find $a$

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

$13 = 4a + b$

Since $b = 1$, we get :

$13 = 4a + 1$

$12 = 4a$

$3 = a$

$\boxed{a = 3}$

3. Conclusion

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

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

Let's check those results :

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

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

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

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

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

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

Done !
_________________

Does it make sense ?

4. Hello, brianfisher1208!

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

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

Um . . . -2 ?

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

We can use any two values from our table.

For example:

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

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

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

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

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

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

5. Interesting! As you can see, Brian, Soroban chose to solve by substitution a system of linear equations in order to find $a$ and $b$, 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