slope, m: = −39. Finally, we plug in the point (5, −29) and the slope m =
−39 to get the equation of the tangent line: y + 29 = −39(x − 5).
- y − 7 = (x − 4)
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 slope by plugging x = 4 into the derivative of the equation y = . First, we
take the derivative: = . Now, we can find the slope, m:
= . Finally, we plug in the point (4, 7) and the slope m = to get the
equation of the tangent line: y − 7 = (x − 4).
- x = ±
The slope of the line y = x is 1, so we want to know where the slope of the tangent line is equal
to 1. We find the slope of the tangent line by taking the derivative: = 6x^2 − 8. Now we set
the derivative equal to 1 : 6x^2 − 8 = 1. If we solve for x, we get x = ±.
- y − 7 = (x − 3)
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 = 3 into the equation y = , and we find the slope
by plugging x = 3 into the derivative of the equation.
First, we find the y-coordinate, y 1 : y = = 7. This means that the line passes through the
point (3, 7).
Next, we take the derivative: = . Now, we can find the
slope, m: = −2. However, this is the slope of the tangent line. The normal line
