Write an equation in slope-intercept form of the line that passes through the given point and is perpendicular to the graph of the given equation.
(-2, 3); y = 12x - 1
To find the equation of a line that is perpendicular to the given line, we need to find the negative reciprocal of the slope of the given line. The given line has a slope of 12, so the negative reciprocal is $\frac{-1}{12}$.
Now, we can use the point-slope form of a linear equation to find the equation of the line. The point-slope form is given by $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is the given point and $m$ is the slope.
Using the point $(-2, 3)$ and the slope $\frac{-1}{12}$, we can substitute into the point-slope form:
$y - 3 = \frac{-1}{12}(x - (-2))$
Simplifying,
$y - 3 = \frac{-1}{12}(x + 2)$
$y - 3 = \frac{-1}{12}x - \frac{1}{6}$
Adding $\frac{1}{6}$ to both sides,
$y = \frac{-1}{12}x + \frac{1}{6} + 3$
$y = \frac{-1}{12}x + \frac{1}{6} + \frac{18}{6}$
$y = \frac{-1}{12}x + \frac{19}{6}$
Therefore, the equation of the line that passes through the point $(-2, 3)$ and is perpendicular to the graph of $y = 12x - 1$ is $y = \frac{-1}{12}x + \frac{19}{6}$.