Question

1. find the slope of the curve at the given point p, and an equation of the tangent line at p.

$y = x^{2}-2x - 3$, $p(2,-3)$

Explanation:

Step1: Find the derivative of the function

The derivative of $y = x^{2}-2x - 3$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ is $y'=\frac{d}{dx}(x^{2}-2x - 3)=2x-2$.

Step2: Calculate the slope at the given point

Substitute $x = 2$ into $y'$. When $x = 2$, $y'(2)=2\times2 - 2=2$. So the slope $m = 2$.

Step3: Use the point - slope form to find the tangent line equation

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(2,-3)$ and $m = 2$.
Substitute these values: $y-(-3)=2(x - 2)$.
Simplify the equation: $y + 3=2x-4$, which gives $y=2x-7$.

Answer:

The slope of the curve at point $P$ is $2$, and the equation of the tangent line at $P$ is $y = 2x-7$.