point (4, 7).
Next, we take the derivative: = . Now, we can find the
slope, m: = −6. However, this is the slope of the tangent line. The normal
line is perpendicular to the tangent line, so its slope will be the negative reciprocal of the
tangent line’s slope. In this case, the slope of the normal line is . Finally, we plug in the
point (4, 7) and the slope m = to get the equation of the normal line: y − 7 = (x − 4).
- y = 0
Remember that the equation of a line through a point (x 1 , y 1 ) with slope m is y m y − y 1 = m(x −
x 1 ). We find the y-coordinate by plugging x = 2 into the equation y = 2x^3 − 3x^2 − 12x + 20, and
we find the slope by plugging x = 2 into the derivative of the equation.
First, we find the y-coordinate, y 1 : y = 2(2)^3 − 3(2)^2 − 12(2) + 20 = 0. This means that the line
passes through the point (2, 0).
Next, we take the derivative: = 6x^2 − 6x − 12. Now, we can find the slope, m: =
6(2)^2 − 6(2) − 12 = 0. Finally, we plug in the point (2, 0) and the slope m = 0 to get the
equation of the tangent line: y − 0 = 0(x − 2) or y = 0.
- y + 29 = −39(x − 5)
Remember that the equation of a line through a point (x 1 , y 1 ) with slope m is y − y 1 = m(x − x 1 ).
We find the y-coordinate by plugging x = 5 into the equation y = , and we find the slope
by plugging x = 5 into the derivative of the equation.
First, we find the y-coordinate, y 1 : y = = -29. This means that the line passes through
the point (5, −29).
Next, we take the derivative: = . Now, we can find the
