Answer:
Step-by-step explanation:
P=(2,-3)
Q=(a,a²-2a-3)
[tex]\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\\[/tex]
Proof:
y'=(x²-2x-3)'=2x-2
y'(2)=2*2-2=2
