~~Find the equation(s) of the tangent line(s) to the graph of the curve

y = x3 − 8x
 through the point

(1, −8)
 not on the graph.
in Calculus Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

y=x³-8x, y'=3x²-8.

The gradient is 3x²-8 at point x. So, at point (p,p³-8p), the equation of the tangent line is y=(3p²-8)x+a. To find a, we plug in the point: p³-8p=(3p²-8)p+a, a=p³-8p-3p³+8p, making a=-2p³, and y=(3p²-8)x-2p³. But this tangent line must also pass through the given point (1,-8), so -8=3p²-8-2p³. So 3p²-2p³=0=p²(3-2p), and p=0 or 3/2. The equations of the tangents are y=-8x (when p=0 at (0,0)) and y=(27/4-8)x-27/4=-5x/4-27/4 (when p=3/2 at (3/2,-69/8)).

The curve is shown in red. The two tangents intersect at (1,-8) and are represented by the blue and green lines.

by Top Rated User (762k points)

Related questions

Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
85,064 questions
90,193 answers
57,138 users