Question

find the point on the parabola y= -x^2+3x+4 where the slope of the tangent is 5

Answer 1

\(\bf \cfrac{d}{dx}[ -x^2+3x+4]={\color{red}{ 5}}\)

so whatever "x" you get will be \(\bf (\square , {\color{red}{ 5}})\)