by plugging x = 3 into the derivative of the equation.
First, we find the y-coordinate, y 1 : y = (3)^3 − 3(3) = 18. This means that the line passes
through the point (3, 18).
Next, we take the derivative: = 3x^2 − 3. Now, we can find the slope, m : = 3(3)^2 − 3
= 24. Finally, we plug in the point (3, 18) and the slope m = 24 to get the equation of the
tangent line: y − 18 = 24(x − 3).
- y − = − (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 = . This means that the line passes through
the point .
Next, we take the derivative: = = . Now, we can find the
slope, m: . Finally, we plug in the point and the slope m = −
to get the equation of the tangent line: y − = − (x − 3).
- 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 y-coordinate by plugging x = 4 into the equation y = , and we find the slope
by plugging x = 4 into the derivative of the equation.
First, we find the y-coordinate, y 1 : y = = 7. This means that the line passes through the
