find the slope of the parabola y equals to e^(-4x) sin6x at the point P(2,4). write an equation for the tangent to the parabola at this point​

Answer :

The equation of the tangent line to the curve is:

[tex]y - 4 = 0.0024(x - 2)[/tex]

Which has a slope of 0.0024.

The tangent line of a curve [tex]f(x)[/tex] at a point [tex](x_0,y_0)[/tex] is given by:

[tex]y - y_0 = f^{\prime}(x_0)(x - x_0)[/tex]

In this problem, the curve is:

[tex]f(x) = e^{-4x}\sin{6x}[/tex]

The derivative is:

[tex]f^{\prime}(x) = -4e^{-4x}\sin{6x} + 6e^{-4x}\cos{6x}[/tex]

Point P(2,4), thus [tex]x_0 = 2, y_0 = 4[/tex].

The slope is:

[tex]f^{\prime}(2) = -4e^{-4(2)}\sin{6(2)} + 6e^{-4(2)}\cos{6(2)} = 0.0024[/tex]

Then, the equation of the line is:

[tex]y - y_0 = f^{\prime}(x_0)(x - x_0)[/tex]

[tex]y - 4 = 0.0024(x - 2)[/tex]

A similar problem is given at https://brainly.com/question/22426360

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