Answer
a: Lowest estimate at $(\frac{3}{2}, -\frac{2}{3})$, and leftmost at $(-1.1,1.1)$
b: Actual lowest at $(1.417, -0.4712)$, and leftmost at $(-1.191, 1.191)$
Work Step by Step
a: With Desmos, we can see that the lowest point on the function is somewhere around $(\frac{3}{2}, -\frac{2}{3})$, and the leftmost point is around $(-1.1,1.1)$
b: b: To find the exact values, we need to find the lowest horizontal asymptote, and thevertical asymptote closest to the left, respectively.
$\frac{dx}{dt}=4t^{3}-2$
$\frac{dy}{dt}=1+4t^{3}$
$\frac{dy}{dx}=\frac{1+4t^{3}}{4t^{3}-2}$
Horizontal Asymptote:
$1+4t^{3}=0$
$4t^{3}=-1$
$t^{3}=-\frac{1}{4}$
$t=-\sqrt[3] \frac{1}{4}$
Plug the values back into x and y.
$x=t^{4}-2t= 1.417$
$y=t+t^{4}= -0.472$
The coordinates of the lowest value are $(1.417,-0.472)$
Vertical Asymptote:
$4t^{3}-2=0$
$4t^{3}=2$
$t^{3}=\frac{2}{4}$
$t=\sqrt[3] \frac{1}{2}$
$x=t^{4}-2t= -1.191$
$y=t+t^{4}= 0.191$
The coordinates of the leftmost value are $(-1.191,1.191)$
