Answer:
The horizontal tangents occur at: (9,-2) and (11,2)
The vertical tangent occurs at (8.75,-1.375)
See attachment
Step-by-step explanation:
The given parametric equations are:
[tex]x=t^2-t+9[/tex] and [tex]y=t^3-3t[/tex]
The slope function is given by:
[tex]\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }[/tex]
[tex]\frac{dy}{dx}=\frac{3t^2-3}{2t-1}[/tex]
The tangent is vertical when [tex]\frac{dx}{dt}=0[/tex]
[tex]\implies 2t-1=0[/tex]
[tex]t=\frac{1}{2}[/tex]
When [tex]t=\frac{1}{2}[/tex], [tex]x=(\frac{1}{2})^2-\frac{1}{2}+9=8.75[/tex], [tex]y=0.5^3-3(0.5)=-1.375[/tex]
The vertical tangent occurs at (8.75,-1.375)
The tangent is horizontal when [tex]\frac{dy}{dt}=0[/tex]
[tex]3t^2-3=0[/tex]
[tex]\implies t=\pm1[/tex]
When t=1, [tex]x=(1)^2-1+9=9[/tex], [tex]y=1^3-3(1)=-2[/tex]
When t=-1, [tex]x=(-1)^2+1+9=11[/tex], [tex]y=(-1)^3-3(-1)=2[/tex]
The horizontal tangents occur at: (9,-2) and (11,2)
