What is the equation of the normal line of f(x)=x+2x−4 at x=3?

At x=3, f(3)=3+23−4=5−1=−5Hence normal is desired at point (3, -5) on the curve.Slope of tangent is given by f'(x)=(x−4)⋅ddx(x+2)−(x+2)⋅ddx(x−4)(x−4)2=(x−4)⋅1−(x+2)⋅1(x−4)2=x−4−x−2(x−4)2=−2(x−4)2Hence slope of tangent at (3, -5) is −2(3−4)2=−2Hence slope of normal would be −1−2=12And equation of the normal line using point slope form is(y−(−5))=12(x−3) or2*(y+5)=(x-3) or x-2y-13=0

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High School
Calculus and Analysis
Calculus 1
Derivatives

Meghan Delacruz

Answered question

2022-04-10

What is the equation of the normal line of

f (x) = \frac{x + 2}{x - 4}

at x=3?

Answer & Explanation

posciad48o

Beginner2022-04-11Added 15 answers

At x=3,

f (3) = \frac{3 + 2}{3 - 4} = \frac{5}{- 1} = - 5

Hence normal is desired at point (3, -5) on the curve.
Slope of tangent is given by f'(x)

= \frac{(x - 4) \cdot \frac{d}{d x} (x + 2) - (x + 2) \cdot \frac{d}{d x} (x - 4)}{{(x - 4)}^{2}}

= \frac{(x - 4) \cdot 1 - (x + 2) \cdot 1}{{(x - 4)}^{2}} = \frac{x - 4 - x - 2}{{(x - 4)}^{2}} = - \frac{2}{{(x - 4)}^{2}}

Hence slope of tangent at (3, -5) is

- \frac{2}{{(3 - 4)}^{2}} = - 2

Hence slope of normal would be

- \frac{1}{- 2} = \frac{1}{2}

And equation of the normal line using point slope form is

(y - (- 5)) = \frac{1}{2} (x - 3)

or
2*(y+5)=(x-3) or x-2y-13=0

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