I have a curve at $c(t) = (-5t^2-3t+4,t^3-9t+5)$ and given a slope for the tangent line of $3$. I would like to find the point $(x,y)$ where this occurs.
What I did is took the derivatives of $x(t)$ and $y(t)$, came up with an equation for the slope of a tangent $y'(t)/x'(t)$ and then set that equal to 3. That gave me $t=-10,0$ which I then plugged back into the original $x(t)$ and $y(t)$ which gave me the points $(4,5)$ and $(-466,905)$. This doesn't seem to be the answer, however, and I'm stumped as to why. I did another similar problem and this method gave me the correct answer. I feel like I might be missing points, but I'm not sure.
Thanks for any help
Presumably your $c'(t)=(-10t-3,3t^2-9)$ and you solved $\frac{3t^2-9}{-10t-3}=3$ The $y$ coordinate at $t=-10$ is $(-10)^3-9(-10)+5=-1000+90+5=-905$