First, we find the derivative of the curve to find the slope of the tangent line at x=1:
y = 2x^3 - x^2 + 3x + 2
y' = 6x^2 - 2x + 3
Evaluate the derivative at x=1:
y'(1) = 6(1)^2 - 2(1) + 3
y'(1) = 6 - 2 + 3
y'(1) = 7
So, the slope of the tangent line at x=1 is 7.
We can use the point-slope form of a line, y - y1 = m(x - x1), to find the equation of the tangent line. Substituting x=1, y=2, and m=7 (the slope we found) into the equation:
y - 2 = 7(x - 1)
y - 2 = 7x - 7
y = 7x - 5
Therefore, the equation of the tangent line to the curve at x=1 is y = 7x - 5.
To find the equation of the normal line, we use the fact that the slopes of a tangent and normal line are negative reciprocals of each other. Therefore, the slope of the normal line is -1/7.
Using the point-slope form of a line again, we have:
y - 2 = (-1/7)(x - 1)
y - 2 = (-1/7)x + 1/7
y = (-1/7)x + 15/7
Therefore, the equation of the normal line to the curve at x=1 is y = (-1/7)x + 15/7.
Find the equation of the tangent and the normal lines to the curves
y=2x^3-x^2+3x+2 at the point x=1
1 answer