1. Write the equation of the line that passes through the negative zero of the polynomial $p(x)=-x^{4}+2 x^{3}+x^{2}-2 x$ and the intersection of its graph with the vertical line that is two units to the left of the y-axis.

(7 points)
1. We set up the equation $-x^{4}+2 x^{3}+x^{2}-2 x=0$, factor out the common factor $-x\left(x^{3}-2 x^{2}-x+2\right)=0$, then factor the quartic polynomial to get $-x(x-2)(x-1)(x+1)=0$. The negative root is $x=-1$. We determine that the vertical line two units to the left of the y-axis is the line $x=-2$. We calculate the intersections of the polynomial graph with this line by evaluating $p(-2)=-24$. We have two points on the desired line: $(-1,0)$ and $(-2,-24)$. We calculate $k=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-24-0}{-2-(-1)}=24$. In $y=k x+n$, we substitute one of the points, for example, $(-1,0)$, and get $y=24 x+24$.

Factoring out the common factor $-x\left(x^{3}-2 x^{2}-x+2\right)=0 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ point Factoring the quartic polynomial $-x(x-2)(x-1)(x+1)=0$ or using Horner's algorithm1* point

![](https://cdn.mathpix.com/cropped/2024_06_07_7bf3e9b12130e254269ag-11.jpg?height=54&width=1639&top_left_y=1321&top_left_x=206)
Determining that $x=-2$ is the vertical line.......................................................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_7bf3e9b12130e254269ag-11.jpg?height=57&width=1639&top_left_y=1419&top_left_x=206)

![](https://cdn.mathpix.com/cropped/2024_06_07_7bf3e9b12130e254269ag-11.jpg?height=54&width=1642&top_left_y=1469&top_left_x=207)

![](https://cdn.mathpix.com/cropped/2024_06_07_7bf3e9b12130e254269ag-11.jpg?height=51&width=1639&top_left_y=1525&top_left_x=206)