[Pre-Calc] Need Help with this Question Please

Thanks Chris,

What graphing program did you use Chris?

I played with this question too and I found it quite interesting. My results were the same as yours ofcourse.  :)

Here are the Desmos graphs, 

https://www.desmos.com/calculator/zcpydgu9sb

$If \quad x=t^2\;\quad y=t+1\qquad \mbox{The restriction is } x\ge0\\ t=\pm\sqrt{x}\qquad t=y-1\\ y-1=\pm \sqrt{x}\\ (y-1)^2=x\\ x=(y-1)^2\\ $

As Chris said this is a sidways parabola opening in the positive direction with a vertex at (0,1)

Lets look at the second one

a) x=sin^2(t), y=sin(t)+1

The first thing I notice is that t is replaced with sin(t)

so we have 

$x=sin^2t \qquad 0\le x\le 1\\ y=sin(t)+1 \qquad 0\le y \le 2\\ \mbox{So the domain and range are different.}\\ But\;\\ sin^{-1}(\pm\sqrt{x})=t \qquad and \qquad sin^{-1}(y-1)=t\\ \therefore\\ sin^{-1}(\pm\sqrt{x})=sin^{-1}(y-1)\\ \pm\sqrt{x}=(y-1)\\ x=(y-1)^2\\ x=(y-1)^2\\ \mbox{So this is exactly the same as the original}\\ \mbox{ relation except it has different a domain and range}$

The first set of equations is

x = t^2   and   y = t + 1

Re-arranging the second one, we have y - 1  = t     and substituting this one into the first one, we have :

x = (y - 1)^2   which is a "sideways" parabola opening to the right with a vertex at (0, 1)

a) x = sin^2t    y = sint + 1

Rearranging the second one again, we have  sint = y - 1    and substituting this one into the first we have

x = (y - 1)^2 

However......this graph is a "restricted" version of the first one......from t = 0 to t = pi/2, the graph traces the same path from the vertex (0,1) to the point (1, 2).....then from t = pi/2 to t =pi.....it "reverses" course from (1,2) back to the vertex......from t = pi to t = 3pi/2, it traces the same path as the original graph from the vertex to the point (1, 0)...lastly, from t = 3pi/2 to 2pi......the graph again reverses course and ends back at the vertex.......this behavior is repeated infinitly

Here's the graph  :

b) x= t^4    y = t^2 + 1

Rearranging the second equation, we have y - 1 = t^2

Substituting this into the first equation we have the original equation, again, x = (y - 1)^2

However, this graph is just the [unbounded] upper half of the parabola since y is never less than 1

For  the interval where t ranges from (-infinity, 0), the graph traces back from the right to the point (0, 1).....when t ranges from (0, infinity), the gragh "reverses" course and traces back to the right

Here's the graph here traced from t = -2 to t = 2 :

Thanks, Melody.....I just graphed these in WolframAlpha and took a  "screen snip"

Thanks Chris :)