Slope-Intercept Form of a Line: Edgenuity Answer

Answer

Change the value of $b$ to $2$.

The function is already in slope-intercept form:

$$h(x)=mx+b=-\frac{2}{3}x+5$$

Here, $m=-\frac{2}{3}$ controls the line's slope and $b=5$ is the $y$-intercept. Moving the entire graph down 3 units subtracts 3 from every output:

$$h(x)-3=-\frac{2}{3}x+5-3=-\frac{2}{3}x+2$$

So the transformed line is

$$h_{new}(x)=-\frac{2}{3}x+2.$$

The slope remains $-\frac{2}{3}$, which means the new line is parallel to the original. Only its $y$-intercept changes, from $(0,5)$ to $(0,2)$.

Why the other choices are incorrect

• Changing $b$ to $-3$ moves the intercept from $5$ to $-3$, a downward shift of 8 units rather than 3.
• Changing $m$ to $-3$ changes the steepness of the line instead of translating it vertically.
• Changing $m$ to $2$ changes both the steepness and direction of the line; it does not produce a 3-unit downward shift.

Quick rule

For any line $y=mx+b$, a vertical shift by $k$ units produces $y=mx+(b+k)$. Keep $m$ unchanged and add the shift to $b$. For a downward shift of 3, $k=-3$, so $b_{new}=5+(-3)=2$.