Find the value of k

**fabxx** · September 19th 2008, 09:09 PM

(page 350 #9)
Sorry for the sloppy graph. The outter line should be a curve concave down. And inside the curve is a triangle.

The question: The figure above shows the graph of y=k-x^2, where k is constant. If the area of triangle ABC is 64, what is the value of k? I don't get why the correct answer is 16. Thanks in advance

**mr fantastic** · September 19th 2008, 09:26 PM

Originally Posted by fabxx

(page 350 #9)
Sorry for the sloppy graph. The outter line should be a curve concave down. And inside the curve is a triangle.

The question: The figure above shows the graph of y=k-x^2, where k is constant. If the area of triangle ABC is 64, what is the value of k? I don't get why the correct answer is 16. Thanks in advance

Are points A, B and C meant to coincide with the vertices of the triangle?

**fabxx** · September 19th 2008, 10:35 PM

yes they are. sorry for the sloppy graph. A, C should be the points for both the parabola and the triangle.

**Moo** · September 19th 2008, 10:44 PM

Okay

Originally Posted by fabxx

yes they are. sorry for the sloppy graph. A, C should be the points for both the parabola and the triangle.

Note that k is positive.
A and C are the x intercepts of the curve y=k-x².
This means that y=0 for these points.
Thus x=+ or - sqrt(k)
The distance AB is then 2sqrt(k).

Can you see what the altitude of the triangle from point B is ? It's the y-ordinate of B, which is the y intercept of the curve. This means that its x is 0. So what is its y ?

Then remember that area of triangle = (1/2)*AB*altitude.

Solve for k !

**mr fantastic** · September 19th 2008, 10:44 PM

Originally Posted by fabxx

(page 350 #9)
Sorry for the sloppy graph. The outter line should be a curve concave down. And inside the curve is a triangle.

The question: The figure above shows the graph of y=k-x^2, where k is constant. If the area of triangle ABC is 64, what is the value of k? I don't get why the correct answer is 16. Thanks in advance

The x-intercepts of the parabola are $x = \pm \sqrt{k}$ and the y-coordinate of the turning point is $y = k$ . So you have a triangle of baselength $2\sqrt{k}$ and height $k$ .

Therefore solve $k \sqrt{k} = 64 \Rightarrow k^{3/2} = 64$ for k.

**fabxx** · September 20th 2008, 01:52 AM

Originally Posted by Moo

Okay
A and B are the x intercepts of the curve y=k-x².

shouldn't the x intercepts of the curve be A and C?

Also another question is why do you make y=0?

Thanks in advance

**fabxx** · September 20th 2008, 02:05 AM

Originally Posted by mr fantastic

Therefore solve $k \sqrt{k} = 64 \Rightarrow k^{3/2} = 64$ for k.

I don't get how you got from $k \sqrt{k} = 64 to \Rightarrow k^{3/2} = 64$ and to the final k=16.
Can you please show me the steps?

Thanks in advance

**Moo** · September 20th 2008, 02:07 AM

Originally Posted by fabxx

shouldn't the x intercepts of the curve be A and C?

Also another question is why do you make y=0?

Thanks in advance

Excuse me, I put B instead of C. Now I've edited.

The x intercepts are the points intercept points of the curve with the x-axis. But any point on the x-axis has a null ordinate. This is why y=0 for these points.

Originally Posted by fabxx

I don't get how you got from $k \sqrt{k} = 64 to \Rightarrow k^{3/2} = 64$ and to the final k=16.
Can you please show me the steps?

Thanks in advance