Question

Find the slope of the tangent at $(1,2)$ on the curve $y=x^2-4x+5$.

Answer 1

Slope of the tangent to a curve at $(x,y)$ is the value of $\frac{dy}{dx}$ at $(x,y)$

$⟹\frac{dy}{dx}=\frac{d(x^2-4x+5)}{dx}⟹\frac{dy}{dx}=2x-4$

At $(1,2),\frac{dy}{dx}=2\left(1\right)-4=-2$

Slope of the tangent $=-2$