Answer
The curve
\[
x=t^2, \quad y=1-3t, \quad z=1+t^2
\] Passes through \((1,4,0)\) and \((9, -8, 28)\), but not through \((4,7,-6)\).
Work Step by Step
We have:
\[
x=t^2, \quad y=1-3t, \quad z=1+t^2
\]
We are going to verify for each point.
Point \((1,4,0)\):
\[
x=t^2=1 \\
y=1-3t=4 \\
z=1+t^3 = 0
\]
Solving for \(t\):
\[
t^2=1 \implies t=\pm1 \\
1-3t=4 \implies t= -1 \\
1+t^3 \implies t=-1
\]
Since \(t=-1\) satisfies all three equations, the curve passes through \((1,4,0)\) at \(t=-1\).
Point \((9, -8, 28)\):
\[
x=t^2=9 \\
y=1-3t=-8 \\
z=1+t^3 = 28
\]
Solving for \(t\):
\[
t^2=9 \implies t= \pm 3 \\
1-3t=-8 \implies t=3 \\
1+t^3=28 \implies t=3
\]
Since \(t=3\) satisfies all three equations, the curve passes through \((9, -8, 28)\) at \(t=3\).
Point \((4,7,-6)\):
\[
x=t^2=4 \\
y=1-3t=7 \\
z= 1+t^3 =-6
\]
Solving for \(t\):
\[
t^2=4 \implies t= \pm 2 \\
1-3t=7 \implies t=-2 \\
1+t^3=-6 \implies t= \sqrt[3]{-7}
\]
We cannot find a single value of \(t\) that solves all 3 equations, therefore the curve does \(\textbf{not}\) pass through \((4,7,-6)\)
