Question: User: y = -2x +3
y= \frac{1}{2}x -\frac{5}{2}
14.
y = 15
Slope =
y -intercept =
17.
Graph shape: a steep positive-sloping line with a plotted point near the upper-left area of the graph; position_hint = bottom-left14.
y = 15
Slope =
y-intercept =
graph shape: a slanted line segment/side runs from upper-left to lower-left, meeting a nearly horizontal top line and a nearly horizontal bottom line, with position_hint=centerThe slope is \frac{-1}{3}. The point (6, 0) lies on
the line.
Graph: coordinate plane with x- and y-axes crossing at the origin; a line is implied by the prompt with slope \frac{-1}{3} passing through (6, 0), position_hint=center
1. **State the problem:**
We are given several linear equations and asked to find the slope and y-intercept for some, and analyze a line with slope $-\frac{1}{3}$ passing through the point $(6,0)$.
2. **Recall the slope-intercept form:**
The equation of a line in slope-intercept form is $$y = mx + b$$ where $m$ is the slope and $b$ is the y-intercept.
3. **Analyze the first equation:**
Given $$y = -2x + 3$$
- Slope $m = -2$
- Y-intercept $b = 3$
4. **Analyze the second equation:**
Given $$y = \frac{1}{2}x - \frac{5}{2}$$
- Slope $m = \frac{1}{2}$
- Y-intercept $b = -\frac{5}{2}$
5. **For the equation $y = 15$:**
This is a horizontal line where $y$ is always 15.
- Slope $m = 0$ (horizontal lines have zero slope)
- Y-intercept $b = 15$
6. **For the line with slope $-\frac{1}{3}$ passing through $(6,0)$:**
Use point-slope form:
$$y - y_1 = m(x - x_1)$$
Substitute $m = -\frac{1}{3}$, $x_1 = 6$, $y_1 = 0$:
$$y - 0 = -\frac{1}{3}(x - 6)$$
Simplify:
$$y = -\frac{1}{3}x + 2$$
So the slope is $-\frac{1}{3}$ and the y-intercept is $2$.
7. **Summary:**
- For $y = -2x + 3$: slope $-2$, y-intercept $3$
- For $y = \frac{1}{2}x - \frac{5}{2}$: slope $\frac{1}{2}$, y-intercept $-\frac{5}{2}$
- For $y = 15$: slope $0$, y-intercept $15$
- For line with slope $-\frac{1}{3}$ through $(6,0)$: equation $y = -\frac{1}{3}x + 2$, slope $-\frac{1}{3}$, y-intercept $2$