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From the given table and options, we can note a linear behavior. Lets find the slope m and the y-intercept b for our table. As we know the line equation is


y=mx+b

where the slope m is given by


m=(y_2-y_1)/(x_2-x_1)

where the values on the right hand side come from 2 points of our table, for instance, we can choose points


\begin{gathered} (x_1,y_1)=(2,32) \\ (x_2,y_2)=(5,50) \end{gathered}

by substtuting these values into the slope formula, we get


m=(50-32)/(5-2)

which gives


\begin{gathered} m=(18)/(3) \\ m=6 \end{gathered}

then, our line equation has the form


y=6x+b

In order to find b, we must substitute one of the 2 choosen points, if we substitute point (2,32), we get


32=6(2)+b

then, we have


\begin{gathered} 32=12+b \\ 32-12=b \\ b=20 \end{gathered}

then, the searched line is


y=6x+20

which corresponds to the last option

User Adrien Castagliola
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