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

The tangent of y=x^3-2x^2+5,(2,5) is y=4x-3

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tangent of y=x^3-2x^2+5,(2,5)

tangent of

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

y = 4 x - 3

Solution steps

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

(2, 5) : m = 4

Find the line with slope m=

4

and passing through

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

y = 4 x - 3

x^{2} y^{^{'}} + 2 x y = ln (x), y (1) = 8

\int \frac{x}{( x + 1 ) ^{2}} d x

\int \frac{x + 3}{x ^{2} + 4} d x

\int \frac{1}{3 sin ( x ) + 4 cos ( x )} d x

\frac{d}{d x} ((5 x^{3} + 4)^{2})

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