Question

graph the line with y-intercept 7 and slope $-\frac{3}{2}$.

Explanation:

Step1: Recall the slope - intercept form

The slope - intercept form of a line is $y = mx + b$, where $m$ is the slope and $b$ is the $y$-intercept. Given that the $y$-intercept $b = 7$ and the slope $m=-\frac{3}{2}$, the equation of the line is $y=-\frac{3}{2}x + 7$.

Step2: Plot the $y$-intercept

The $y$-intercept is the point where the line crosses the $y$-axis. When $x = 0$, $y=7$. So we plot the point $(0,7)$ on the coordinate plane.

Step3: Use the slope to find another point

The slope $m = \frac{\text{rise}}{\text{run}}=-\frac{3}{2}$. The negative sign means we can go down 3 units (since rise is - 3) and then run 2 units to the right (run is 2), or we can go up 3 units and run 2 units to the left.
Starting from the point $(0,7)$, if we go down 3 units (so $y=7 - 3=4$) and run 2 units to the right (so $x = 0+2 = 2$), we get the point $(2,4)$.
We can also go up 3 units ( $y=7 + 3 = 10$, but 10 is not in our visible graph range) or run 2 units to the left from $(0,7)$: $x=0 - 2=- 2$ and $y=7+3 = 10$ (also out of range) or $x = 0-2=-2$ and $y=7 - 3=4$? Wait, no. Wait, slope is $\frac{\text{rise}}{\text{run}}=-\frac{3}{2}=\frac{3}{-2}$. So another way is rise = 3, run=-2. So from $(0,7)$, if we rise 3 ( $y = 7+3 = 10$) and run - 2 ( $x=0-2=-2$), we get $(-2,10)$ (out of range). If we rise - 3 ( $y=7 - 3 = 4$) and run 2 ( $x = 0 + 2=2$), we get $(2,4)$ (in range). If we rise - 3 and run - 2 ( $x=0-2=-2$, $y=7 - 3=4$? No, $\frac{-3}{-2}=\frac{3}{2}$, which is positive. Wait, no. The slope is $-\frac{3}{2}$, so $\text{rise}=-3$, $\text{run}=2$ or $\text{rise}=3$, $\text{run}=-2$.
So from $(0,7)$, using $\text{rise}=-3$, $\text{run}=2$: new $x=0 + 2=2$, new $y=7-3 = 4$, so point $(2,4)$.
Using $\text{rise}=3$, $\text{run}=-2$: new $x=0-2=-2$, new $y=7 + 3=10$ (not in the given graph's $y$-range which goes from - 8 to 8). So we use $(2,4)$ and $(0,7)$.

Step4: Draw the line

Draw a straight line passing through the points $(0,7)$ and $(2,4)$ (and extend it in both directions).

(Note: Since the problem is about graphing, the key steps are identifying the equation, plotting the $y$-intercept, using the slope to find another point, and then drawing the line. If we were to describe the graph, the line will have a negative slope, crossing the $y$-axis at $(0,7)$ and passing through $(2,4)$ etc.)

Answer:

To graph the line:

1. Plot the $y$-intercept point $(0,7)$ (where the line crosses the $y$-axis).
2. Use the slope $-\frac{3}{2}$ (which means for every 2 units you move to the right along the $x$-axis, you move down 3 units along the $y$-axis, or for every 2 units left, you move up 3 units). From $(0,7)$, moving 2 units right (to $x = 2$) and 3 units down (to $y=7 - 3 = 4$) gives the point $(2,4)$.
3. Draw a straight line through $(0,7)$ and $(2,4)$ (and extend it in both directions) to represent the line $y=-\frac{3}{2}x + 7$.