desperately need help on this calc problem! anyone?

Question

Let f(x)=12-x^2 for x is greater than or equal to 0 and f(x) is greater than or equal to 0.

a. The line tangent to the graph of f at the point (k, f(k) intercepts the x-axis at x=4. What is the value of k?

b. An isosceles triangle whose base is the interval from (0,0) to (c, 0) has its vertex on the graph f. For what value of c does the triangle have maximum area? Justify your answer...

simple explanations would be very very helpful! ;)

Answer 1

y = 12 - x^2

y' = - 2x is the first derivative (slope of line tangent)

y - y1 = m (x - x1) is the equation of the line with slope m and passing the point (x1, y1)

y - 0 = - 2x (x - 4) is the equation of the line tangent passing

the point (4,0)

then, y = - 2x^2 + 8x

therefore, we can do:

12 - x^2 = - 2x^2 + 8x

The solution of this quadratic equation the point where the parabole intercepts the line tangent, and solving the equation, you find 2 solutions:

x = 2 & x = 6 where x = 2 is the valid one. Plug this value on the original f(x), and you find y = 8

Then k = 2 and f(k) = 8

Graph both, the parabole and tangent for clarification...

Answer 2

Rewrite equation as f(x)= -x`2+12
This is a parabola identical to f(x)=x`2 in shape and size except that the "-" indicates it opens downward and the 12 means the vertex is (0,12). I have to give the tangent question more thought.