Question

For parts (a) and (b) below, guess what might be Latifah's rule for the function. in each case, if n is the input and L(n) is the answer, express L(n) in terms of n. a. A possible rule could be L(n)= ______

Answer 1

a)

To obtain the rule we notice that for each unit we change in n the change in L(n) is 4 and that this change is constant (the same) this means that this is a line. A line can be represented as:

$L(n)=\frac{L(n_2)-L(n_1)}{n_2-n_1_{}}(n-n_1)+L(n_1)$

(This formula is the same as:

$y=m(x-x_1)+y_1$

but with the variables rename for this specific problem)

Using the first two rows we have:

$\begin{gathered} L(n)=\frac{17-13}{5-4}(n-4)+13 \\ L(n)=\frac{4}{1}(n-4)+13 \\ L(n)=4(n-4)+13 \\ L(n)=4n-16+13 \\ L(n)=4n-3 \end{gathered}$

Therefore the function is:

$L(n)=4n-3$

b)

For the second table we notice that the change is not constant, this means that this is not a line. Then we need to try with another type of function. The second function we usually try is a quadratic function of the form:

$L(n)=an^2+bn+c$

We need to determine the value of the constants a, b and c. To do this we use the first three rows of the table to get three equations. For n=2 we have:

$8=4a+2b+c$

For n=5 we have:

$29=25a+5b+c$

For n=9 we have:

$85=81a+9b+c$

Then we have the system of equations:

$\begin{gathered} 4a+2b+c=8 \\ 25a+5b+c=29 \\ 81a+9b+c=85 \end{gathered}$

To solve this system we can solve the first equation for c, then we have:

$c=8-4a-2b$

Plugging this in the second we have:

$\begin{gathered} 25a+5b+8-4a-2b=29 \\ 21a+3b=21 \end{gathered}$

And in the third equation we have:

$\begin{gathered} 81a+9b+8-4a-2b=85 \\ 77a+7b=77 \end{gathered}$

Then we have the system of equations:

$\begin{gathered} 21a+3b=21 \\ 77a+7b=77 \end{gathered}$

Now, to solve it we rewrite the equations as:

$\begin{gathered} 7a+b=7 \\ 11a+b=11 \end{gathered}$

we got this by dividing the first one by 3 and the second by 7. Now if we substract the first equation from the second we have:

$\begin{gathered} 7a-11a+b-b=7-11 \\ -4a=-4 \\ a=\frac{-4}{-4} \\ a=1 \end{gathered}$

Once we know the value of a we plug it in one of the equations of the second system to get b, then:

$\begin{gathered} 7(1)+b=7 \\ 7+b=7 \\ b=7-7 \\ b=0 \end{gathered}$

Now we can get the value of c:

$\begin{gathered} c=8-4(1)-2(0) \\ c=8-4-0 \\ c=4 \end{gathered}$

Now we plug the values of a, b and c in the function we propose. Then the function we are looking for is:

$L(n)=n^2+4$