Question

A curve C is defined by the parametric equations x=t^2-4t+1 and y=t^3 what is the equation of the line tangent to the point (-2, 27)?

Answer 1

the derivative of t^2-4t+1 = 2t-4. Plugging in -2 gives 2(-2)-4 which is 0. therefore y=mx+c
y=0x+c
y=27
assuming the point (-2,27) exists

Answer 2

okay i see thanks

Answer 3

so the equation would be y=27 ?

Answer 4

Yes :)

Answer 5

thanks!