SOLUTION: a right triangle has one vertex on the graph of y = x^2 at (x,y),another at the origin, and the third on the positive y- axis at (0,y). express the area A of the triangle as a func

Question 194333: a right triangle has one vertex on the graph of y = x^2 at (x,y),another at the origin, and the third on the positive y- axis at (0,y). express the area A of the triangle as a function of x. for what value of x will be the area of the triangle be equal to 30 square units?
please help me to solve!!!!
please i really don't know how to solve!!!
please.....help
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
First, let's draw a picture:

Now since the point (x,y) is "x" units to the right and "y" units up, this means that the base of the triangle is "x" units long and the height is "y" units high. Now let's label the drawing:

But remember, we let . So replace "y" with to get

Now focusing on just the triangle, we get the following

So the base is "x" and the height is

Now recall (or look up), the area of any triangle is

Start with the given formula

Plug in and

Multiply

So the area is

"for what value of x will be the area of the triangle be equal to 30 square units?"

Start with the given equation.

Plug in

Multiply both sides by 2.

Multiply

Rearrange the equation

Take the cube root of both sides.

Approximate the right side with a calculator.

So the answer is approximately

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