Question: 2. Find the slope-intercept form of the line that contains the point $(-2, 4)$ and is perpendicular to $3x + 2y = 15$.\nm = \frac{y_2 - y_1}{x_2 - x_1}\ny = mx + b\n3. Find the slope-intercept form of the line that contains the points $(6, 2)$ and $(-3, -4)$.\nm = \frac{-4 + -2}{6 + 3} = \frac{-6}{9} = -\frac{2}{3}\ny = -\frac{2}{3} x +\nGraph the piecewise defined function (on graph paper. Attach to back of test.)\nf(x) = \begin{cases} -x + 4, & x < 1 \\ 2x + 1, & x \geq 1 \end{cases}
1. **Problem:** Find the slope-intercept form of the line that contains the point $(-2, 4)$ and is perpendicular to the line given by the equation $$3x + 2y = 15$$.
2. **Step 1: Find the slope of the given line.**
Rewrite the equation in slope-intercept form $y = mx + b$:
$$3x + 2y = 15$$
$$2y = -3x + 15$$
$$y = -\frac{3}{2}x + \frac{15}{2}$$
So, the slope of the given line is $$m = -\frac{3}{2}$$.
3. **Step 2: Find the slope of the perpendicular line.**
The slope of a line perpendicular to another is the negative reciprocal of the original slope:
$$m_{\perp} = -\frac{1}{m} = -\frac{1}{-\frac{3}{2}} = \frac{2}{3}$$
4. **Step 3: Use the point-slope form to find the equation of the perpendicular line.**
The point-slope form is:
$$y - y_1 = m(x - x_1)$$
Using point $(-2, 4)$ and slope $\frac{2}{3}$:
$$y - 4 = \frac{2}{3}(x - (-2))$$
$$y - 4 = \frac{2}{3}(x + 2)$$
5. **Step 4: Simplify to slope-intercept form.**
$$y - 4 = \frac{2}{3}x + \frac{4}{3}$$
$$y = \frac{2}{3}x + \frac{4}{3} + 4$$
$$y = \frac{2}{3}x + \frac{4}{3} + \frac{12}{3}$$
$$y = \frac{2}{3}x + \frac{16}{3}$$
**Final answer:** $$y = \frac{2}{3}x + \frac{16}{3}$$