# Find the slope of the tangent to the curve

Question:

Find the slope of the tangent to the curve $y=x^{3}-3 x+2$ at the point whose $x$-coordinate is $3 .$

Solution:

The given curve is $y=x^{3}-3 x+2$.

$\therefore \frac{d y}{d x}=3 x^{2}-3$

The slope of the tangent to a curve at $\left(x_{0}, y_{0}\right)$ is $\left.\frac{d y}{d x}\right\rfloor_{\left(x_{0}, y_{0}\right)}$.

Hence, the slope of the tangent at the point where the x-coordinate is 3 is given by,

$\left.\left.\frac{d y}{d x}\right]_{x=3}=3 x^{2}-3\right]_{x=3}=3(3)^{2}-3=27-3=24$