What is the tangent of y=3x^3-x^2+4,(2,24) ?

The tangent of y=3x^3-x^2+4,(2,24) is y=32x-40

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tangent of y=3x^3-x^2+4,(2,24)

tangent of

y = 3 x^{3} - x^{2} + 4, (2, 24)

y = 32 x - 40

Solution steps

Find the slope of

y = 3 x^{3} - x^{2} + 4 : \frac{d y}{d x} = 9 x^{2} - 2 x

Compute the slope of

y = 3 x^{3} - x^{2} + 4

(2, 24) : m = 32

Find the line with slope m=

32

and passing through

(2, 24) : y = 32 x - 40

y = 32 x - 40

h \to + 0 lim (\frac{8 ^{h} - 1}{h})

\frac{x ^{2} e ^{x}}{x ^{2} + e ^{x}}

y^{^{''}} + 25 y = 0

\frac{d}{d x} (x sec (\frac{1}{x}))

\int \frac{x ^{3}}{4 x ^{2} + 16} d x

What is the tangent of y=3x^3-x^2+4,(2,24) ?
The tangent of y=3x^3-x^2+4,(2,24) is y=32x-40