## Precalculus (6th Edition)

$y=\frac{1}{4}x+\frac{13}{4}$
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 $\color{blue}{y=\frac{1}{4}x+\frac{13}{4}}$.