[Math] f(x)=(x-a)(x-b)(x-c) help

calculus

I have been assigned a homework assignment and my proffesor always gives very vague questions and no guidance.

f(x)=(x-a)(x-b)(x-c), assume a,b and c to be constants.

  1. Find two values of x that make f(x)=0

    I know that in this form, x=a.x=b and x=c are all zeroes of this function. I still don't get what is being asked of me. This only complicates everything because without part 1 I can't do the rest.

  2. Find the average of the two values of x you found in part one. Which point on the graph of f has this x value?

  3. Find the equation for the tangent line of f at the point you found in part 2.

  4. Find all od the points where the tangent line from part three intersects the graph y=f(x), where f is defined above.

Best Answer

1)You're correct. $a,b,c$ are the zeroes. 2)Hint;The average of two of those x's is $\frac{a+b}{2}$ and the point is $M(\frac{a+b}{2}, f(\frac{a+b}{2}))$

3)Hint; What does the derivative have to do with tangent lines?

4)Hint;You can find them by setting the tangent from 3) equal to $f(x)$. You'll notice something interesting too.

Your attitude towards the teacher is ridiculous though.

Related Question