Find an equation of the tangent line to the curve at the given point (2, -4) y = 4x - 3x²

Solution:

Given, the expression is y = 4x - 3x²

We have to find an equation of the tangent line to the curve at the given point (2, -4).

Slope of the curve is found by differentiating y = 4x - 3x²

dy/dx = 4 - 6x

At point (2, -4)

dy/dx = 4 - 6(2)

= 4 - 12

= -8

dy/dx means the slope m

So, m = -8

The equation of the line in point-slope form is given by

\((y-y_{1})=m(x-x_{1})\)

Here, (x₁, y₁) = (2, -4) and m = -8.

\((y-(-4))=-8(x-2)\\(y+4)=-8(x-2)\)

y + 4 = -8x + 16

y = -8x + 16 - 4

y = -8x + 12

Therefore, the equation of the tangent line is y = -8x + 12.

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Find an equation of the tangent line to the curve at the given point (2, -4) y = 4x - 3x²

Summary:

An equation of the tangent line to the curve y = 4x - 3x² at the given point (2, -4) is y = -8x + 12.