Answer
A parabola that opens in the negative $y$ direction on the plane $x = -3$, with a vertex at $(-3, 1, 0)$
Work Step by Step
Step 1
From a vector-valued function: \[ \mathbf{r} = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k} \] the parametric equations can be obtained: \[ \begin{align*} x &= x(t) \\ y &= y(t) \\ z &= z(t) \end{align*} \] The graph can be identified by remembering certain forms, using trigonometric relationships, converting to Cartesian form, or by plotting some points and drawing a smooth curve through them.
Step 2
From: \[ \mathbf{r} = -3t\mathbf{i} + (1 - t^2)\mathbf{j} + t\mathbf{k} \] we have parametric equations: \[ \begin{align*} x &= -3 \\ y &= 1 - t^2 \\ z &= t \end{align*} \] With the constant $x = -3$, the action takes place on the plane $x = -3$. The equation $y = 1 - t^2$ represents a parabola that opens in the negative $y$ direction.
Step 3
The vertex is located where the derivative is $0$: \[ \frac{dy}{dt} = -2t = 0 \implies t = 0 \] The vertex point is then $(x, y, z) = (-3, 1, 0)$. Result A parabola that opens in the negative $y$ direction on the plane $x = -3$, with a vertex at $(-3, 1, 0)$.
