Answer
The tangent line has a slope of $\frac{1}{2}$ at the point $(4,0)$
Work Step by Step
$x = 3t^2+1$
$y = t^3-1$
We can find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^2}{6t} = \frac{t}{2}$
We can find $t$ when $\frac{dy}{dx} = \frac{1}{2}$:
$\frac{t}{2} = \frac{1}{2}$
$t = 1$
When $t = 1$:
$x = 3(1)^2+1 = 4$
$y = (1)^3-1 = 0$
The tangent line has a slope of $\frac{1}{2}$ at the point $(4,0)$
