But what is that last equation you wrote?

Also, the math script isn't necessary unless you

like to type even more to get it to look cool.

As long as your understood.

But [m a t h]\frac{3}{4}[/m a t h] makes a 3/4 fraction.

You can google on LaTeX Math to learn it.

\int is integrate

_{x=0} makes x=0 a subscript, the underscore does that.

^{I'm tiny north east} makes writing a superscript.

I'm going to bed now, it's 1:45 am here. Bye.

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would use the product rule.

f'g+g'f

+ J*(-1/(x+1)^2)

I don't know how to use th math script..

(x^2+2x+2)/(x+1)^2

]]>And noted the minimum area was 4 at x=2

I'm rusty at calculus, but I want to learn it again years later.

So this is a product, so there is a rule for that.

What is the derivative of this?

dJ

I'll show my teacher your method.

]]>Add 2 to both sides to solve for Yintercept.

Multiply both sides of equation by Xintercept so as to

solve for the area; you can divide by 2, if you want to

be exact.

What I'm most curious is if this congruency is true at all points.

]]>X intercept = 2

Y intercept = 4

Call x intercept J.

Here is the equation for the area.

Just find the J above one that makes the expression the smallest:

I arrived at this by drawing a picture and setting two

similar triangles proportions equal.

They are:

Sorry explanation so poor.]]>

y = mx + b

We know this has to pass through the point (x, y) = (1, 2), so lets plug those in:

2 = m + b, or, m = 2 - b

That didn't seem to get us very far, did it? But at least we have a relationship for m and b, this might come in handy later.

So lets go back. What we want is a general equation for the area of the triangle. Well, we know of the formula 1/2 * base * height. So lets try to find those variables.

The height of the triangle is going to be the y-intercept, b. Easy enough.

The base of the triangle is going to be the x intercept. Since y = mx + b, and y must be 0 (definition of the x-intercept) 0 = mx + b. We want to solve this for x, so that would be x = -b / m.

So the area of the triangle is 1/2 * b * -b/m, or -b^2 / 2m. But wait, isn't this going to be negative? Negative area? Nope, remember the line that we are drawing has a negative slope, so m is negative, making -b^2 / m positive.

So we want to find the least area of the function -b^2 / 2m. Huh, two variables, that's going to be pretty tricky without multi variable calculus. But wait, doesn't m = 2 - b? Told you that would come in handy. So -b^2 / 2 * (2 - b) is the area, or -b^2 / (4 - 2b).

Try to find the minimum for that function. This will tell you what b is, then you can find m because m = 2 - b.

Edit:

And for extra credit, what kind of triangle does this make?

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