	AllExperts > Calculus
	Search

About Scotto
Expertise
Any kind of mathematics (calculus, analysis, game theory, linear approximation, finite differences, linear regression, linear programming, numerical analysis, probability, statistics, etc.). I also have answered some questions in Physics (mass, momentum, falling bodies), Chemistry (charge, reactions, symbols, molecules), and Biology.

Experience
Experience in the area: I have tutored students in all areas of mathematics for over 20 years. Education/Credentials: BSand MS in Mathematics from Oregon State University, where I completed sophomore course in Physics and Chemistry. I received both degrees with high honors. Awards and Honors: I have passed Actuarial tests 100, 110, and 135.

Publications
Maybe not a publication, but I have respond to well oveer 3000 questions on the PC. That's around 2,000 in basic math and 1,000 in advanced math.

Education/Credentials
I aquired well over 40 hours of upper division courses. This was well over the number that were required. I graduated with honors in both my BS and MS degree from Oregon State University. I was allowed to jump into a few junior level courses my sophomore year.

Awards and Honors
I have been nominated as the expert of the month several times. All of my scores right now are at least a 9.8 average (out of 10).

Past/Present Clients
My past clients have been students at OSU, students at the college in South Seattle, referals from a company, friends and aquantenances, people from my church, and people like you.

You are here: Experts > Teens > Homework/Study Tips > Calculus > calculus

Calculus - calculus

Expert: Scotto - 11/4/2009

Question
Find the equation of the tangent line to the curve f(x)=x^3 that is parallel to the line x-(1/12)y+3=0.

Answer
The equation of the line should be put in the form where the slope can be read.
That is, y = mx + b, and m is the slope.

Adding y/12 to both sides gives x + 3 = y/12. Flipping the equation gives y/12 = x + 3.
Multiplying both sides by 12 gives y = 12x + 36. This means the slope is 12.

Now if the curve that touches on this tangent line has slope 12,
that would mean the derivative at that point would be 12.
Since the derivative is f'(x) = 3x², and that is 12, we take 12 = 3x².
Division of both sides by 3 gives 4 = x².
The answer to that is ±2. This means that there are two tangent lines.

The y value for 2 is 8 and the y value for -2 is -8. The points are (8,2) and (-8,-2).
The line at (8,2) is y-2 = 12(x-8) => y-2 = 12x - 96 => y = 12x-94.
The line at (-8,-2) is y+2 = 12(x+8) => y+2 = 12x + 96 => y = 12x + 94.

Thus, there are two lines that meet the requirements, and I found both of them.

Add to this Answer Ask a Question