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

The tangent of y=x^3-2x^2+5,\at x=2 is y=4x-3

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tangent of y=x^3-2x^2+5,\at x=2

tangent of

y = x^{3} - 2 x^{2} + 5, at x = 2

y = 4 x - 3

Solution steps

Find the tangent point:

(2, 5)

Find the slope of

y = x^{3} - 2 x^{2} + 5 : \frac{d y}{d x} = 3 x^{2} - 4 x

Compute the slope of

y = x^{3} - 2 x^{2} + 5

x = 2 : m = 4

Find the line with slope m=

4

and passing through

(2, 5) : y = 4 x - 3

y = 4 x - 3

ln (π)

x \to - 0.1 lim (\frac{sin ( x )}{x})

n = 1 \sum \infty \frac{1}{7 + e ^{- n}}

n \to \infty lim (3^{n} 7^{- n})

x \to 9 - lim (x^{2} + 5 x + 7)

What is the tangent of y=x^3-2x^2+5,\at x=2 ?
The tangent of y=x^3-2x^2+5,\at x=2 is y=4x-3