Hello there. To solve this question, we'll have to remember some properties about orientation of parametric curves and how to determine their graphs.
First, given the parametric equations:
[tex]\begin{gathered} x=t^2+2 \\ y=t^3-4t \end{gathered}[/tex]
We can solve the first equation for t, such that:
[tex]\begin{gathered} t^2+2=x \\ t^2=x-2 \\ t=\pm\sqrt[]{x-2} \end{gathered}[/tex]
Since t is between - 3 and 3, we have to analyze which values of x are possible using the end points:
(-3)² + 2 = 3² + 2 = 9 + 2 = 11
So as you can see, the values of x are
[tex]2\le x\le11[/tex]
Plugging the first solution into the equation for y, we have
[tex]\begin{gathered} _{}y=(\sqrt[]{x-2})^3-4\sqrt[]{x-2}_{} \\ y=(x-6)\sqrt[]{x-2} \end{gathered}[/tex]
And as x is between 2 and 11, we get that y is between
[tex]-15\le y\le15[/tex]
Sketching this curve, we have
With the other solution, we get
[tex]\begin{gathered} _{}y=(-\sqrt[]{x-2})^3-4\cdot(-\sqrt[]{x-2}) \\ y=-(x-2)\sqrt[]{x-2}+4\sqrt[]{x-2} \\ y=(6-x)\sqrt[]{x-2} \end{gathered}[/tex]
The two curves together looks like
Finally, we have to determine the orientation
For this, let's take a point in the first solution, say we take t = 1 then t = 2
Plugging it into the equations, we get
[tex]\begin{gathered} x=1^2+2=1+2=3 \\ y=1^3-4\cdot1=-3 \end{gathered}[/tex]
Now with t = 2
[tex]\begin{gathered} x=2^2+2=4+2=6 \\ y=2^3-4\cdot2=0 \end{gathered}[/tex]
So as t increses, we go from:
So this is the orientation of the curve and it is the answer contained in the fourth option.
