(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
Okay
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 !
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.
and remember a rule of exponents :