Question

Find an equation of the tangent line to the curve y = 8 - 2x - 3x^2 at the point (1,3).

A) y = -7x + 10
B) y = -5x + 6
C) y = -4x + 5
D) y = -2x + 3

Answer 1

To find the equation of the tangent line to y = 8 - 2x - 3x^2 at (1,3), we need to find the slope at that point and use the point-slope form. The equation of the tangent line is y = -8x + 11.

Step-by-step explanation:

To find the equation of the tangent line to the curve y = 8 - 2x - 3x2 at the point (1,3), we need to find the slope of the curve at that point and then use the point-slope form of a linear equation to find the equation of the tangent line.

Step 1: Take the derivative of the curve y = 8 - 2x - 3x2 to find the slope function: y' = -2 - 6x

Step 2: Evaluate the slope function at x = 1 to find the slope at the point (1,3): y' = -2 - 6(1) = -8

Step 3: Use the point-slope form of a linear equation, y - y1 = m(x - x1), with m = -8 and (x1, y1) = (1,3), to find the equation of the tangent line: y - 3 = -8(x - 1) = -8x + 8

The equation of the tangent line to the curve y = 8 - 2x - 3x2 at the point (1,3) is y = -8x + 11.