Question

Determine the tangent equation for y=2-3x^2+4x at x=2.

Answer 1

To determine the tangent equation at x = 2, we first need to find the derivative of the function y = 2 - 3x^2 + 4x.

Taking the derivative with respect to x, we get:

dy/dx = -6x + 4

Now, we can substitute the value x = 2 into the derivative to find the slope of the tangent line:

dy/dx = -6(2) + 4
= -12 + 4
= -8

Therefore, the slope of the tangent line at x = 2 is -8.

To find the equation, we need to use the point-slope form of a linear equation, where the equation is given by y - y1 = m(x - x1), with (x1, y1) being a point on the line and m being the slope.

Using the point (2, y), where y = 2 - 3(2)^2 + 4(2) = 2 - 12 + 8 = -2, the equation becomes:

y - (-2) = -8(x - 2)

Simplifying further:

y + 2 = -8x + 16

Rearranging to slope-intercept form, we get:

y = -8x + 14

Therefore, the tangent equation for y = 2 - 3x^2 + 4x at x = 2 is y = -8x + 14.