Answer
See below:
Work Step by Step
We have to graph the provided equations; also, we have to eliminate the common variable t and make a single equation in x and y.
The provided parametric equation in y is $ y=2t $. Computing the value of t from this equation, we get:
$\begin{align}
& y=2t \\
& t=\frac{y}{2} \\
\end{align}$
By putting this value of t in the equation of x, we get:
$\begin{align}
& x={{t}^{2}}+t+1 \\
& x={{\left( \frac{y}{2} \right)}^{2}}+\frac{y}{2}+1 \\
\end{align}$
And solving this equation further, we get:
$\begin{align}
& x={{\left( \frac{y}{2} \right)}^{2}}+\frac{y}{2}+1 \\
& x=\frac{{{y}^{2}}}{4}+\frac{y}{2}+1 \\
& x=\frac{{{y}^{2}}+2y+4}{4}
\end{align}$
And completing the square, we get:
$\begin{align}
& x=\frac{{{y}^{2}}+2y+4}{4} \\
& x=\frac{{{y}^{2}}+2y+1+3}{4} \\
& x=\frac{{{\left( y+1 \right)}^{2}}+3}{4} \\
\end{align}$
Then multiplying both sides by 4 and then subtracting 3 from both sides, we get:
$\begin{align}
& 4x={{\left( y+1 \right)}^{2}}+3 \\
& 4x-3={{\left( y+1 \right)}^{2}} \\
\end{align}$
Then writing the equation in the form of ${{y}^{2}}=4ax $, we get:
$\begin{align}
& {{\left( y+1 \right)}^{2}}=4x-3 \\
& {{\left( y+1 \right)}^{2}}=4\left( x-\frac{3}{4} \right) \\
\end{align}$
And the above equation is of the form of a parabola, having vertex $\left( \frac{3}{4},-1 \right)$.
We see from the equations that the domain and range are:
Domain: $[\dfrac{3}{4},\infty]$
Range: $[-\infty,\infty]$
