A right triangle has one vertex on the graph of y=16-x^2, x>0, at (x,y), another at the origin, and the 3rd pn the positive x-axis t (x, 0). Express the area A of the triangle as a function of x.

Question

help

Answer 1

For a triangle, A = 1/2bh

In this case "x" is the base and "16-x²" is the height so just plug them in.

Answer 2

right triangle has one vertex on the graph of y =16-x^2 with x>0

another vertex is the origin

3rd vertex is (x,0)

describe area as a function of x. A = f(x) = ?

base = x

height = 16-x^2

A = bh/2 = x(16-x^2)/2 = 8x - (x^3)/2

only problem is while (0, 16) is the top vertex on the line y=16-x^2, x is not >0 as required

so real answer is No solution. DNE

or alternatively

(epislon, 16) is the top vertex, then you can get

A = 8x -(x^3)/2

epsilon = 0 + an infinitesmal = 1/infinity