Chapter 6.3, Problem 16E

Solution

To eliminate: The parameter and identify the graph. Use the graph to support your answer.

The general equation is $y=x^2-3$ and the graph is shown in figure $\left(1\right)$ .

Given information:

The equations are $x=t$ and $y=t^2-3$ .

Calculation:

Substitute $x$ for $t$ to the equation $y=t^2-3$ .

  $y=x^2-3$

This is a parabola opening upward with vertex at $\left(0,-3\right)$ from the vertex form $y=a{\left(x-h\right)}^2+k$ . This is validated by the graph obtained from the parametric equations.

  

Figure (1)

Therefore, the general equation is $y=x^2-3$ and the graph is shown in figure $\left(1\right)$ .