Question

(a) Find the slope of the curve y= x^2 - 2x - 3 at the point ​P(2​,​ -3) by finding the limit of the secant slopes through point P.
(b) Find an equation of the tangent line to the curve at P(2, -3)

Answer 1

Answer:

Step-by-step explanation:

P=(2,-3)

Q=(a,a²-2a-3)

$\Delta y=a^2-2a-3-(-3)=a^2-2a\\\Delta x=a-2\\\\\displaystyle \lim_{a \to 2} \dfrac{a^2-2a-3-(-3)}{a-2} \\\\=\lim_{a \to 2} \dfrac{a^2-2a}{a-2} \\\\=\lim_{a \to 2} \dfrac{a(a-2)}{a-2} \\\\=\lim_{a \to 2} \dfrac{a}{1} \\\\=\lim_{a \to 2} a \\\\=2$

Proof:

y'=(x²-2x-3)'=2x-2

y'(2)=2*2-2=2