Differentiation proof

Question

Find the co-ordinates of the point on a curve $y=x^2+3x-1$ at which it is parallel to the line $ y=5x-1?$

unsure how to solve this

Answer 1

Assuming you want tangent line of the $y=x^2+3x-1$ to be parallel to $y=5x-1$, solution is:

Find derivative of the first line $$y=x^2+3x-1 \Rightarrow y'=2x+3.$$ For the lines to be parallel, their derivatives have to be the same $$y=5x-1 \Rightarrow y'=5$$ $$2x+3=5 \Rightarrow x=1.$$ We can find the coordinates if we plug the $x=1$ into our first line $$y=1+3-1=3.$$ Coordinates are $(1,3)$