Answer
$y=\frac{1}{4}x+\frac{13}{4}$
Work Step by Step
RECALL:
(1) The slope-intercept form of a line's equation is $y=mx+b$ where $m$=slope and $(0, b)$ is the line's y-intercept.
(2) The slope $m$ of a line that contains the points $(x_1, y_2)$ and $(x_2, y_2)$ is given by the formula $m=\dfrac{y_2-y_1}{x_2-x_1}$..
Solve for the slope using the formula in (2) above to obtain:
$m=\dfrac{4-3}{3-(-1)}=\dfrac{1}{3+1}=\dfrac{1}{4}$
Thus, the tentative equation of the line is $y=\frac{1}{4}x+b$
To find the value of $b$, substitute the x and y values of one of the two given points to obtain:
$y=\frac{1}{4}x+b
\\4=\frac{1}{4}(3)+b
\\4=\frac{3}{4}+b
\\4-\frac{3}{4}=\frac{3}{4}+b-\frac{3}{4}
\\\frac{16}{4}-\frac{3}{4}=b
\\\frac{13}{4}=b$
Therefore, the equation of the line is $\color{blue}{y=\frac{1}{4}x+\frac{13}{4}}$.
