What is the tangent of f(x)=x^2-9x+18,4,\at x=3 ?

The tangent of f(x)=x^2-9x+18,4,\at x=3 is y=-3x+9

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tangent of f(x)=x^2-9x+18,4,\at x=3

tangent of

f (x) = x^{2} - 9 x + 18, 4, at x = 3

y = - 3 x + 9

Solution steps

Find the tangent point:

(3, 0)

Find the slope of

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

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 = - 3

Find the line with slope m=

- 3

and passing through

(3, 0) : y = - 3 x + 9

y = - 3 x + 9

\frac{d}{d x} (\frac{x}{( 3 x - 8 ) ^{8}})

t an g e n t f (x) = \frac{3}{x + 3}, at x = - \frac{1}{3}

x \to 0 lim (cos (x) \cdot x)

\frac{d ^{4}}{d x ^{4}} (2 ln (x))

in v erse l a pl a ce \frac{20}{s ( s + 30 )}

What is the tangent of f(x)=x^2-9x+18,4,\at x=3 ?
The tangent of f(x)=x^2-9x+18,4,\at x=3 is y=-3x+9