For a function  to be continuous at a given point , it must meet the following two conditions:

1.) The point  must exist, and

2.) .

Examining the above graph,  is continuous at every possible value of  except for , . Thus,  is continuous on the interval $\displaystyle (-\infty,0)\cup(0,\infty)$.

For a function  to be continuous at a given point , it must meet the following two conditions:

1.) The point  must exist, and

2.) .

Examining the above graph,  is continuous at every possible value of  except for , $\displaystyle x=-3$. Thus,  is continuous on the interval $\displaystyle (-\infty,-3)\cup(-3,\infty)$.

For a function to be continuous at a given point , it must meet the following two conditions:

1.) The point must exist, and

2.) .

Examining the above graph, is continuous at every possible value of except for $\displaystyle x\leq0$. Thus, is continuous on the interval $\displaystyle (0,\infty)$.

$\displaystyle x = t^{2} + 2t + 1$

$\displaystyle x = \left ( t + 1\right )^{2}$

$\displaystyle \pm \sqrt{x} = t + 1$

So 

$\displaystyle \sqrt{x} = t + 1$ or $\displaystyle -\sqrt{x} = t + 1$

We are restricting $\displaystyle t$ to values on $\displaystyle \left [ -1, 1\right ]$, so $\displaystyle t + 1$ is nonnegative; we choose 

$\displaystyle \sqrt{x} = t + 1$.

Also,

$\displaystyle y = t^{2} - 2t + 1$

$\displaystyle y= \left ( t - 1\right )^{2}$

$\displaystyle \sqrt{y} = \pm \left ( t - 1\right )$

So 

$\displaystyle \sqrt{y} = t- 1$ or $\displaystyle -\sqrt{y} = t- 1$

We are restricting $\displaystyle t$ to values on $\displaystyle \left [ -1, 1\right ]$, so $\displaystyle t - 1$ is nonpositive; we choose

$\displaystyle -\sqrt{y} = t- 1$

or equivalently,

$\displaystyle \sqrt{y} = -t+ 1$

to make $\displaystyle t - 1$ nonpositive.

Then,

$\displaystyle \sqrt{x}+ \sqrt{y} = (t + 1) + (-t + 1)$

and 

$\displaystyle \sqrt{x}+ \sqrt{y} = 2$

$\displaystyle \sqrt{y} = 2 - \sqrt{x}$

$\displaystyle y =\left ( 2 - \sqrt{x} \right )^2$

$\displaystyle y = 4 - 4\sqrt{x} + \left (\sqrt{x} \right )^{2}$

$\displaystyle y = x + 4 - 4\sqrt{x}$

Rewrite $\displaystyle x$ using the double-angle formula:

$\displaystyle x = \cos 2t =1 - 2\sin ^{2} t = 1 - 2y^{2}$

Then 

$\displaystyle x = 1 - 2y^{2}$

$\displaystyle x + 2y^{2} = 1$

which is the correct choice.

$\displaystyle x = 3t$, so 

$\displaystyle x^{2} = (3t)^{2} = 9t^{2} = \frac{9}{2} \cdot 2t^{2} = \frac{9}{2} y$.

$\displaystyle \frac{9}{2} y= x^{2}$, so

$\displaystyle \frac{2}{9}\cdot \frac{9}{2} y = \frac{2}{9}x^{2}$

$\displaystyle y = \frac{2}{9}x^{2}$

$\displaystyle y = 4t+2 = 4t - 2 + 4 = 2 (2t - 1)+ 4 = 2x + 4$,

so the Cartesian equation is 

$\displaystyle y = 2x + 4$.

$\displaystyle x = \ln (t+1)$

so 

$\displaystyle e ^{x} = \left [ \ln (t+1) \right ] ^{x}$

$\displaystyle e ^{x} = t + 1$

$\displaystyle y = t^2 + 2t + 2 = t^2 + 2t + 1 + 1 = (t + 1)^{2}+ 1 = \left (e^{x} \right )^{2} + 1 = e^{2x} + 1$

Therefore the Cartesian equation is  $\displaystyle y = e^{2x} + 1$.

$\displaystyle y = \sinh t = \frac{e ^{2t} -1}{2e ^{t}} = \frac{\left (e ^{ t} \right) ^{2} -1}{2e ^{t}}$

$\displaystyle e^{t} = 10 ^{\log e \cdot t} =\left ( 10 ^{ t} \right )^{\log e} = x ^{\log e}$, so

$\displaystyle y = \frac{\left ( x ^{\log e} \right) ^{2} -1}{2 x ^{\log e}}= \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}$

This makes the Cartesian equation

$\displaystyle y = \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}$.

In order to solve this, we must isolate $\displaystyle \small t$ in both equations. 

$\displaystyle \small \small x=3t+4\rightarrow t=(x-4)/3$ and 

$\displaystyle \small y=1/t\rightarrow t=1/y$.

Now we can set the right side of those two equations equal to each other since they both equal $\displaystyle \small t$.

 $\displaystyle \small (x-4)/3=1/y$.

By multiplying both sides by $\displaystyle \small 3y/(x-4)$, we get $\displaystyle \small y=3/(x-4)$, which is our equation in rectangular form.