What is the tangent of f(x)=2x^3-x^2+3,(2,15) ?

The tangent of f(x)=2x^3-x^2+3,(2,15) is y=20x-25

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tangent of f(x)=2x^3-x^2+3,(2,15)

tangent of

f (x) = 2 x^{3} - x^{2} + 3, (2, 15)

y = 20 x - 25

Solution steps

Find the slope of

f (x) = 2 x^{3} - x^{2} + 3 : \frac{df ( x )}{d x} = 6 x^{2} - 2 x

Compute the slope of

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

(2, 15) : m = 20

Find the line with slope m=

20

and passing through

(2, 15) : y = 20 x - 25

y = 20 x - 25

n = 1 \sum \infty \frac{x ^{n}}{n !}

\int \frac{30}{1 - cos ( 5 x )} d x

\int \frac{x ^{6}}{x ^{14} - 1} d x

(15 y^{2} - x y) d x + x^{2} d y = 0

\int_{0}^{1} 3 d x

What is the tangent of f(x)=2x^3-x^2+3,(2,15) ?
The tangent of f(x)=2x^3-x^2+3,(2,15) is y=20x-25