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}})

\frac{4 s}{s ^{2} + 16}

f (x) = \frac{3}{x + 3}, at x = - \frac{1}{3}

y = x^{2}, y = 8 - 2 x

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

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