What is the tangent of f(x)=x^3-5x^2+7,\at x=1 ?

The tangent of f(x)=x^3-5x^2+7,\at x=1 is y=-7x+10

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tangent of f(x)=x^3-5x^2+7,\at x=1

tangent of

f (x) = x^{3} - 5 x^{2} + 7, at x = 1

y = - 7 x + 10

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

f (x) = x^{3} - 5 x^{2} + 7 : \frac{df ( x )}{d x} = 3 x^{2} - 10 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 = - 7

Find the line with slope m=

- 7

and passing through

(1, 3) : y = - 7 x + 10

y = - 7 x + 10

What is the tangent of f(x)=x^3-5x^2+7,\at x=1 ?
The tangent of f(x)=x^3-5x^2+7,\at x=1 is y=-7x+10