What is the tangent of f(x)=x^2-9x+17,\at x=4 ?

The tangent of f(x)=x^2-9x+17,\at x=4 is y=-x+1

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tangent of f(x)=x^2-9x+17,\at x=4

tangent of

f (x) = x^{2} - 9 x + 17, at x = 4

y = - x + 1

Solution steps

Find the tangent point:

(4, - 3)

Find the slope of

f (x) = x^{2} - 9 x + 17 : \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 = - 1

Find the line with slope m=

- 1

and passing through

(4, - 3) : y = - x + 1

y = - x + 1

What is the tangent of f(x)=x^2-9x+17,\at x=4 ?
The tangent of f(x)=x^2-9x+17,\at x=4 is y=-x+1