Find the point on the graph of $g(x) = 2x^2 + 3x + 1$ at which the normal line is parallel to the line with equation $x - y = 2$

Question

I suspect I should use implicit differentiation at some point to solve this problem but I'm somewhat clueless on how to approach this problem. Would appreciate tips and directions

Answer 1

notice, the slope of the tangent to the curve: $g(x)=2x^2+3x+1$ is given as $$g'(x)=4x+3$$ hence, the slope of the normal line at some point $(x, y)$ is $$=-\frac{1}{g'(x)}=-\frac{1}{4x+3}$$ The above line is parallel to the line: $x-y=2$ (slope$=1$) hence comparing the slopes one should get $$-\frac{1}{4x+3}=1$$ $$4x+3=-1\implies x=-1$$ setting $x=-1$ in the equation of curve one should get $$g(-1)=2(-1)^2+3(-1)+1=0$$ hence, the required point on the curve is $\color{red}{(-1, 0)}$