Question

3. *h(4) = -6 and g(-2) = 6. g(x) is perpendicular to h(x) and g(1) = 0. a. Find g(x). b. Find h(x).

Answer 1

Let's find g(x) using the given points:

`\begin{gathered} g(-2)=6 \\ g(1)=0 \\ so\colon \\ (x1,y1)=(-2,6) \\ (x2,y2)=(1,0) \end{gathered}`

Let's find the slope:

`m1=(y2-y1)/(x2-x1)=(0-6)/(1-(-2))=(-6)/(3)=-2`

Using the point-slope equation:

`\begin{gathered} y-y1=m1(x-x2) \\ y-6=-2(x-(-2)) \\ y-6=-2x-4 \\ y=-2x+2 \\ g(x)=-2x+2 \end{gathered}`

Since g(x) and h(x) are parallel, we can conclude:

`\begin{gathered} m1\cdot m2=-1 \\ -2\cdot m2=-1 \\ m2=(1)/(2) \end{gathered}`

Using the given point for h(x):

`\begin{gathered} h(4)=-6 \\ so\colon \\ (x1,y1)=(4,-6) \end{gathered}`

Using the point-slope equation again:

`\begin{gathered} y-y1=m(x-x1) \\ y-(-6)=(1)/(2)(x-4) \\ y+6=(1)/(2)(x-4) \\ y+6=(1)/(2)x-2 \\ y=(1)/(2)x-8 \\ h(x)=(1)/(2)x-8 \end{gathered}`