What is the tangent of f(x)=1-9x^2,\at x=3 ?

The tangent of f(x)=1-9x^2,\at x=3 is y=-54x+82

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tangent of f(x)=1-9x^2,\at x=3

tangent of

f (x) = 1 - 9 x^{2}, at x = 3

y = - 54 x + 82

Solution steps

Find the tangent point:

(3, - 80)

Find the slope of

f (x) = 1 - 9 x^{2} : \frac{df ( x )}{d x} = - 18 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 54

Find the line with slope m=

- 54

and passing through

(3, - 80) : y = - 54 x + 82

y = - 54 x + 82

n = 0 \sum \infty (- \frac{5}{4})^{n}

\frac{d v}{d t} = 9.8 - 0.6

\frac{d}{d x} (x - \frac{1}{2})

\frac{d}{d x} (\frac{2 - x ^{3}}{2 + x ^{3}})

\int_{1}^{4} 2 π x (- \frac{3}{2} x + 6) d x

What is the tangent of f(x)=1-9x^2,\at x=3 ?
The tangent of f(x)=1-9x^2,\at x=3 is y=-54x+82