A right triangle has one vertex on the graph of y = 9 - x2, x 0, and (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.

A(x)=

Question

A right triangle has one vertex on the graph of y = 9 - x2, x > 0, and (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.

A(x)=

myininaya · Answer

i believe 27/2 if i didn't misunderstand

anonymous · Answer

27/2?

anonymous · Answer

im under the impression they want a function based on x.

myininaya · Answer

let me read it again
9x/2

myininaya · Answer

i assumed at first (x,0) touched the parabola

anonymous · Answer

yes they want a function based on x and ill try 9x/2

anonymous · Answer

this is what im getting >.<

anonymous · Answer

okay, ill try it. one sec.

myininaya · Answer

i thought the vertex was touching one point of the triangle

anonymous · Answer

yep that was the answer (: thanks! i got a couple more word problems if you'll keep an eye on the thread for me!

anonymous · Answer

one point of the triangle is on the graph, one is at the origin, and one is somewhere on the x axis. now i dont think my formula is right >.<

anonymous · Answer

.....or maybe it is, whatever >.<

myininaya · Answer

well gj

anonymous · Answer

medals for everyone >.>

myininaya · Answer

okay i misread it

myininaya · Answer

it  didn't say anything about the vertex of the parabola

anonymous · Answer

it happens, no worries :) if i had a nickel for every time i misread something, i would have $124.65

myininaya · Answer

lol

anonymous · Answer

Now that im thinking about it....my drawing doesnt make any sense.  That formula i give only works is x <= 3.  if its outside that parabola my function will give a negative value >.<