Question

which function can be used to model the data in this table?
\begin{tabular}{|c|c|} hline $x$ & $f(x)$ \\ hline $-4$ & $-2$ \\ hline $2$ & $2.5$ \\ hline $12$ & $10$ \\ hline end{tabular}
a $f(x)=\frac{3}{2}x + 4$
b $f(x)=\frac{4}{3}x - 1$
c $f(x)=\frac{3}{4}x - 1$
d $f(x)=\frac{3}{4}x + 1$

Explanation:

Step1: Test option A with \( x = -4 \)

Substitute \( x = -4 \) into \( f(x)=\frac{3}{2}x + 4 \):
\( f(-4)=\frac{3}{2}(-4)+4=-6 + 4=-2 \) (matches the first point). Now test \( x = 2 \):
\( f(2)=\frac{3}{2}(2)+4=3 + 4=7
eq2.5 \) (does not match). So A is incorrect.

Step2: Test option B with \( x = -4 \)

Substitute \( x = -4 \) into \( f(x)=\frac{4}{3}x - 1 \):
\( f(-4)=\frac{4}{3}(-4)-1=-\frac{16}{3}-1=-\frac{19}{3}
eq -2 \) (does not match). So B is incorrect.

Step3: Test option C with \( x = -4 \)

Substitute \( x = -4 \) into \( f(x)=\frac{3}{4}x - 1 \):
\( f(-4)=\frac{3}{4}(-4)-1=-3 - 1=-4
eq -2 \) (does not match). So C is incorrect.

Step4: Test option D with \( x = -4 \)

Substitute \( x = -4 \) into \( f(x)=\frac{3}{4}x + 1 \):
\( f(-4)=\frac{3}{4}(-4)+1=-3 + 1=-2 \) (matches). Test \( x = 2 \):
\( f(2)=\frac{3}{4}(2)+1=\frac{3}{2}+1=2.5 \) (matches). Test \( x = 12 \):
\( f(12)=\frac{3}{4}(12)+1=9 + 1=10 \) (matches all points).

Answer:

D. \( f(x)=\frac{3}{4}x + 1 \)