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Milton Shepard

Answered question

2022-04-14

How do you find the equation of the tangent line to the curve

f (x) = x^{2} + 2 x

; at x=3, x=5?

davelucasnp0j

Beginner2022-04-15Added 16 answers

Slope of tangent at

x = x_{0}

on the curve y=f(x) is the value of f'(x) at

x = x_{0}

i.e.

x = x_{0}

Here we have

f (x) = x^{2} + 2 x

, hence

f^{'} (x) = \frac{d f}{d x} = 2 x + 2

Hence slope of tangent at x=3 is f'(3)=2*3+2=8
and slope of tangent at x=5 is f'(5)=2*5+2=12
Note that when x=3,

f (3) = 3^{2} + 2 \cdot 3 = 15

hence tangent at x=3 asses through (3, 15) and has slope 8 and hence equation is y-15=8(x-3) i.e.
8x-y=9.
and when x=5,

f (3) = 5^{2} + 2 \cdot 5 = 35

hence tangent at x=5 asses through (5,35) and has slope 12 and hence equation is y-35=12(x-5) i.e.
12x-y=25

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