Area of a triange as a function?

Question

A right triangle has one vertex on the graph y= 9 - x^2, x>0 at (x y) another at the origin and the third on the positive x axis at (x 0). Express the area A of the triangle as a function of x.

I said it was A(x) = 1/2 (w *x)

am I right? Please help explain thanks!!

Answer 1

yeeeehaw's answer and explanation are correct until x = 3, at which time the height starts to go negative. For x >3, the height of the triangle is therefore -y and the area can be given as

A(x) = 1/2 * b * h

= 1/2 * x * -(9 - x^2)

= 1/2 * (x^3 - 9x)

= 1/2 * x^3 - 9/2 * x if x> 3

and

A(x) = 9/2 * x - 1/2 * x^3 if 0 < x <= 3.

Answer 2

A = x*y/2 =x*(9-x^2)/2 = (9*x - x^3)/2

When x=1, A=4
x=2, A=5
x=3, A=0

Domain 0=

Answer 3

A = 9x/2 = 9*3/2=27/2 = 13.5 units ^2
The point (x,0) is (3,0) and the point on y=9-x^2 is (0,9).
The above answerers are forgetting that the triangle must be a right triangle. The only right trianle possible is where the coordinate axes are the right angle.

Answer 4

The area will be (1/2)bh

Base is from 0 to x, so it is x

height is from 0 to y, so it is y = 9 - x^2

Area = (1/2)x(9 - x^2)
= (9/2)x - (1/2)x^3

Answer 5

close, it's

A(x) = 1/2( |y|*x), where |y| is the absolute value of y,
because y can be negative when x>3

Answer 6

A(x) = 1/2(Y*X)
so A(X) = 4.5X - 0.5X^3