What is the tangent of f(x)=x^2-8x+9,\at x=3 ?

The tangent of f(x)=x^2-8x+9,\at x=3 is y=-2x

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tangent of f(x)=x^2-8x+9,\at x=3

tangent of

f (x) = x^{2} - 8 x + 9, at x = 3

y = - 2 x

Solution steps

Find the tangent point:

(3, - 6)

Find the slope of

f (x) = x^{2} - 8 x + 9 : \frac{df ( x )}{d x} = 2 x - 8

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 = - 2

Find the line with slope m=

- 2

and passing through

(3, - 6) : y = - 2 x

y = - 2 x

\int_{0}^{1} \frac{2}{1 + t ^{2}} d t

\int \frac{12}{1 - cos ( 4 x )} d x

n = 0 \sum \infty \frac{1}{6}

\int - \frac{3}{2}

f (x) = 2 x^{2}

What is the tangent of f(x)=x^2-8x+9,\at x=3 ?
The tangent of f(x)=x^2-8x+9,\at x=3 is y=-2x