Find all points (if any) of horizontal and vertical tangency to the curve. Use a graphing utility to confirm your results. (If an answer does not exist, enter DNE.) x=1-t , y=t^2

Horizontal tangent
(x, y)=
Vertical tangent
(x, y)=

Horizontal tangent

(x, y) = (1, 0)

Vertical tangent

(x, y) = DNE

Step-by-step explanation:

The equation for the slope (m) of the tangent line at any point of a parametric curve is:

Where and are the first derivatives of the horizontal and vertical components of the parametric curves. Now, the first derivatives are now obtained:

and

The equation of the slope is:

As resulting expression is a linear function, there are no discontinuities and for that reason there are no vertical tangents. However, there is one horizontal tangent, which is:

The point associated with the horizontal tangent is:

The answer is:

Horizontal tangent

(x, y) = (1, 0)

Vertical tangent

(x, y) = DNE

the reason why your question keeps getting deleted because this is not a school related question : )

i will answer your question though.

i never tried church's