What is the tangent of y=x^3+4x-3,(1,2) ?

The tangent of y=x^3+4x-3,(1,2) is y=7x-5

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tangent of y=x^3+4x-3,(1,2)

tangent of

y = x^{3} + 4 x - 3, (1, 2)

y = 7 x - 5

Solution steps

Find the slope of

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

Compute the slope of

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

(1, 2) : m = 7

Find the line with slope m=

7

and passing through

(1, 2) : y = 7 x - 5

y = 7 x - 5

4 e^{a r c c o s (4 x)}

\int 3 x^{3} d x

f (x) = ln (x^{2} - 2 x + 1), at x = 2

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

g (x) = 7 - x

What is the tangent of y=x^3+4x-3,(1,2) ?
The tangent of y=x^3+4x-3,(1,2) is y=7x-5