<s>[INST] Tangent line parallel to another line At what point of the parabola $y=x^2-3x-5$ is the tangent line parallel to $3x-y=2$? Find its equation.
I don't know what the slope of the tangent line will be. Is it the negative reciprocal? 
 [/INST] Edit: since the tangent is parallel to the given line: $3x-y=2$  hence the slope of tangent line to the parabola is $\frac{-3}{-1}=3$ 
Let the equation of the tangent be $y=3x+c$ 
Now, solving the equation of the tangent line: $y=3x+c$ & the parabola: $y=x^2-3x-5$ by substituting $y=3x+c$ as follows $$3x+c=x^2-3x-5$$ $$\implies x^2-6x-(c+5)=0\tag 1$$ For tangency we have the following condition $$ \text{determinant},\ B^2-4AC=0$$ $$\implies (-6)^2-4(1)(-(c+5))$$ $$\implies c=\frac{-56}{4}=-14$$ Hence, setting the value of $c=-14$ we get $$x^2-6x-(-14+5)=0$$ $$\implies x^2-6x+9=0$$ $$\implies (x-3)^2=0\implies x=3$$ Now, setting the value of $x=3$ in the equation of parabola as follows $$y=(3)^2-3(3)-5=-5$$ Hence, the point of tangency is $\color{blue}{(3, -5)}$ 
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