The slope of the tangent to the curve x= t square+3t-8,y=2t square-2t-5 at the point(2,-1) is?

Question

It is related to application of derivatives. That is concept of slopes tangents.

Answer 1

The parametric equations are:

x=t^2+3t-8, dx/dt=2t+3, and

y=2t^2-2t-5, dy/dt=4t-2.

dy/dx=dy/dt÷dx/dt=(4t-2)/(2t+3).

The point (2,-1) is (x,y) so:

2=t^2+3t-8 and -1=2t^2-2t-5,

t^2+3t-10=0=(t+5)(t-2), 2t^2-2t-4=0=2(t^2-t-2)=2(t-2)(t+1). The common factor is t-2=0, so t=2.

This is the value of t which produces the coordinates (x,y)=(2,-1) and defines the tangent at (2,-1), so we substitute t=2 in dy/dx=6/7, the slope of the tangent.