The curve given by the parametric equations;
x=36−t^2, y=t^3−25t
is symmetric about the x-axis. (if t gives us the point (x, y), then −t will give (x,−y) ).
a. at which x value is the tangent to this curve horizontal?
b. at which t value is the tangent to this curve vertical?

x=36, t = ±5/sqrt 3

Step-by-step explanation:

we are given a curve reprsented by parametric equations.

Also given that this is symmetric about x axis.

For finding a tangent we first find slope i.e. dy/dx

If dy/dx has dy/dt in numerator and dx/dt in the denominator

Hence for a tangent to be horizontal slope =0

i.e.dy/dt =0

Similarly for a tangent to be vertical slope=undefined

Or dx/dt =0

a)

when x= 36, the tangent is horizontal

b)

For these t values tangent is vertical.

it has the same slope and a different y-intercept

hope this : )