application of MVT to parabolas

[seanavok]

Consider the parabola y=f(x)=ax^2+bx+c. By the mean value theorem, for any two numbers x1 and x2 there is a number x3 in the interval (x1,x2) such that the tangent to the parabola at the point (x3,f(x3)) is parallel to the secant line joining the points (x1,f(x1)) and (x2,f(x2)). Show that x3=(x1+x2)/2

need help! don't know where to start.


thanks,

Sean

 

[Romsek]

Consider the parabola y=f(x)=ax^2+bx+c. By the mean value theorem, for any two numbers x1 and x2 there is a number x3 in the interval (x1,x2) such that the tangent to the parabola at the point (x3,f(x3)) is parallel to the secant line joining the points (x1,f(x1)) and (x2,f(x2)). Show that x3=(x1+x2)/2

need help! don't know where to start.


thanks,

Sean

start by computing the slope of that secant line joining (x1,f(x1) and (x2,f(x2)).

Then figure out the general expression for the tangent to your parabola at a point x. Figure out from that expression at what point x the slope to the parabola is the same as you found for your secant line.

Give it a shot.

 

[seanavok]

I'm able to do all that, but not sure how to show that it's related to midpoint.

 

[stapel]

need help! don't know where to start.

I'm able to do all that....

Oh. So you did know how to start...? :?

How about, rather than our wasting your time re-creating what you've already done, instead you reply showing your work and reasoning so far. Then we'll be able to tell where you're really having difficulty, and can reply with useful assistance. Thank you!

 

[seanavok]

After he explained to me where to start I was able to calculate the slope of the secant line,
(f(x2)-f(x1))/(x2-x1)
and the slope of the tangent line to any point on the parabola f(x)=ax^2+bx+c,
d^2y/dx^2=2ax+b, but I am not sure where to go from here.

 

[stapel]

...I am not sure where to go from here.

Try here:

Then figure out the general expression for the tangent to your parabola at a point x.