Mathematics: (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)

(a) Find the slope of the curve y= x^2 - 2x -

Ответ:

sindy35111

16.10.2021

Mathematics

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

(a) Find the slope of the curve y= x^2 - 2x - 3 at the point P(2, -3) by finding the limit of the se

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Ответ:

jwyapo4

14.01.2023

Mathematics

x-int=5 ; y-int=-3

Step-by-step explanation:

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(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)

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