Tangent slope

I just had a four-hour-straight math test, and I found one of the questions rather interesting, at least interesting enough to share with you. I found a pretty great solution to it, and I'm interested to see what you guys can come up with.

Determine (and prove) whether or not a secant of a second-degree function on the form (y=ax^2+bx), which always goes through origo and one more point on the curve, has the same slope as a tangent of the curve which tangents the curve on the arithmetic mean value of the two x values of the secant. The curve always has a vertex (extreme point) in origo.

what level math are you in? Also the last word is unfamiliar to me, origo.

what level math are you in? Also the last word is unfamiliar to me, origo.

Eh, a bunch of different classes. For this particular problem, I'd personally recommend calculus solutions. And origo is another word for origin, or the point where the x and y axes intersect.

Bleh, not very scientific, but then again, I haven't worked with tangents and secants in quite some time now (at least 5 years...)

Well, firstly, the only time a vertex of y=ax^2+bx, will occur at the origin is when b=0, since the bx term is a linear shift. Now we are working with y=ax^2.

For arguements sake, set a=1, arbitrarily, since it is the easiest to work with.

This means, we can take the two easiest points, (0,0) and (1,1). The arithmetic mean of the x terms is (1+0)/2 = 1/2.

So at x=1/2, y=(1/2)^2=1/4.

The slope of the secant line for points (0,0) and (1,1) is m=1.

Here comes the "not-so-scientific" part of the proof.

For a slope of the tangent line at point (1/2, 1/4) to be m=1, a quick check can be used. If x=0.5000001, then at a slope of 1, y=0.2500001. However, when you plug x=0.5000001 into y=x^2, you get y=0.25000010000001, which is not what we expected. The same happens for any number you plug in, showing that the slope must be 1, which is equal to the secant's slope. The only reason I used such a small number is to show that even at infinitly small increments, the points from a slope = 1 will never touch the y=x^2 curve again.

However, another quick check would be to use a slope = 1.000001. Again, if you plug in numbers for x, you will see that at another point, which I don't feel like finding, there will be a corresponding y point that matches the curve, which shows the line is not tangent.

If you really wanted, you could keep the "a" term and you will see that the coefficients of the "a" term won't line up with a slope of the secant, which is good, but with any other slope, the coefficients will match, showing that the line is not tangent. Bleh.

Like I said, had I been thinking, this is much easier, but I now take for granted what the meaning of particular methods are, such as is the actual proof for solving this problem.

I'll just put my general solution I used on the test.

First, I specified a variable, z, to take the place of x to avoid confusion when comparing the curve to axes. Since the secant goes through (0, 0) and "another point on the curve", I used the point (z, (az^2+bz)). And as Titus said, b has to be zero, or the curve's extreme point won't go through the origin.

Then, I simply used the dy/dx definition to find the slope between the two points:

((az^2+bz)-0) / (z-0) = (az^2+bz) / z = (z(az+b))/z = az+b

To compare the secant to the tangent, I just derived the original curve to find the slope of the tangent at any given point for the same variables:

y=ax^2+bx
y'=2ax+b

Since the x value of the tangent is supposed to be the arithmetic mean value, (z+0)/2 = z/2

So just cram the x co-ordinate into the derivate function:

y'(z/2) = 2a*(z/2)+b = az+b

And there's my "proof" for why the slopes should always be the same for the secant and tangent, if real numbers are being used.

Nevermind.