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

The tangent of f(x)=-5x^2+x,\at x=3 is y=-29x+45

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

tangent of

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

y = - 29 x + 45

Solution steps

Find the tangent point:

(3, - 42)

Find the slope of

f (x) = - 5 x^{2} + x : \frac{df ( x )}{d x} = - 10 x + 1

Compute the slope of

f (x) = - 5 x^{2} + x

x = 3 : m = - 29

Find the line with slope m=

- 29

and passing through

(3, - 42) : y = - 29 x + 45

y = - 29 x + 45

\int_{0}^{1} (x (x + 1)) d x

- \frac{2}{5 x}

f (x) = 3 x + 6 x, (1, 9)

\int_{5}^{\infty} e^{- \frac{x}{3}} d x

\int \frac{2 x}{9 - x ^{2}} d x

What is the tangent of f(x)=-5x^2+x,\at x=3 ?
The tangent of f(x)=-5x^2+x,\at x=3 is y=-29x+45