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**tommyjay117** Thanks, that's a big help. I always forget to substiture the variables from the other formulas that I already know. This is another I was having trouble with, any ideas?

Let f(x)= 12-x² for x ≥ 0 and f(x) ≥ 0

a) The line tangent to the graph of f at the point (k, f(k)) intercepts the x-axis at 4 = x. 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 of f. For what value of c does the triangle have maximum area? Justify your answer.

I know for a that the slope of the line tangent to f is f'(x) = 2x. So the line is y = 2x(x-k) + f(k). it intercepts at (4,0) so 0 = 2(4)(4-k) + 0 => 0 = 32-8k => k = 4. Right? I don't know what to do for b though.