What is the tangent of f(x)=(9-6x)^2,(1,9) ?

The tangent of f(x)=(9-6x)^2,(1,9) is y=-36x+45

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tangent of f(x)=(9-6x)^2,(1,9)

tangent of

f (x) = (9 - 6 x)^{2}, (1, 9)

y = - 36 x + 45

Solution steps

Find the slope of

f (x) = (9 - 6 x)^{2} : \frac{df ( x )}{d x} = - 12 (9 - 6 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 = - 36

Find the line with slope m=

- 36

and passing through

(1, 9) : y = - 36 x + 45

y = - 36 x + 45

\frac{1}{s ^{2} ( 1 + 7 s )}

\frac{d}{d x} (x (1 - \frac{x}{10}))

\frac{d}{d x} (5 x cos (x^{2}))

x \to + 0 lim (\frac{5}{x ^{2} + x} - \frac{5}{x})

ln (\frac{x - 1}{x + 1})

What is the tangent of f(x)=(9-6x)^2,(1,9) ?
The tangent of f(x)=(9-6x)^2,(1,9) is y=-36x+45