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

User Mejan
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Final answer:

First, the derivative of the function y = 4x - 3x² is calculated to find the slope of the tangent line at x = 2. Then, by applying the point-slope form of a line, the equation of the tangent line at point (2, -4) is derived as y = -8x + 12.

Step-by-step explanation:

To find an equation of a tangent line to the curve y = 4x - 3x² at the point (2,-4), we first need to determine the slope of the tangent line. The slope of the curve at any point is given by the derivative of the function. So, we differentiate y with respect to x to get dy/dx = 4 - 6x. Evaluating the derivative at x = 2 gives us dy/dx = 4 - 6(2) = -8, which is the slope of the tangent line at the point (2, -4).

Now, we use the point-slope form to write the equation of the tangent line. The point-slope form is given by y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point on the line. Substituting m = -8 and the coordinates of the point (2, -4), we get the equation of the tangent line as y + 4 = -8(x - 2).

Simplifying this, we get y = -8x + 16 - 4, which further simplifies to y = -8x + 12 as the final equation of the tangent line.

User Findango
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