CPhill

I think you left out part of your question........

So, we will use the change of base rule here....we have

(log 32768) / (log 8) =5

This is a 45 degree in the 3rd quadrant, and the sine of a 45 degree angle is 1/√2 or √2/2. But the sine is negative in this quadrant, so we have -√2/2

Let's find the slope, first = change in y / change in x

Note that from (-4,2) to (6,6),

y changes by 4, and x changes by 10, so the slope is 4/10 = 2/5

So using one of the points - I'll use (6,6) - we have, writing in "point-slope" form:

y - 6 = (2/5)(x-6)

y - 6 = (2/5)x - 12/5       Add 6 to both sides

y = (2/5)x + 18/5           And that's in the form..... y = mx + b

You should verify that both points "check" in this equation........

Let's try this again.......we have

125 / [(5/6) ((15-(2^3 + 3/4)*2))]    Working inside the parentheses in the denominator

125 / [(5/6) ((15-(8 + 3/4)*2))]  =

125/[(5/6) ((15-(35/4)*2))] =

125/[(5/6) (15- (35/2))] =

125/[(5/6) (-5/2)] =

125 / (-25/12) =

-125*(12/25) = (-5 * 12)

-60

OK......what's the question??

I assume that you have this  x√(9-x) = x*(9-x)^(1/2)

Taking the first derivative, we have

(9-x)^(1/2) - (1/2)x(9-x)^(-1/2)     Setting to 0 and factoring, we have

(9-x)^(-1/2) [ (9-x) - (1/2)x]  = 0

[9 - (3/2)x] = 0

9 = (3/2)x

x = 6

We can show that this is a max by taking the second derivative =

[3(x-12)] / [4(9-x)^(3/2)]

And plugging in the critical point x = 6, we see that the numerator < 0 and the denominator > 0  = negative, so this tells us that  (6, 6√3) is the maximum real point on the graph.

zegoes is correct.....your function is linear.....it can be written in the form y= mx + b, i.e.,

y = x + 9

At least one of the "x's" in your original function would have to be squared in order to be quadratic.....

Anonymous is almost correct.

-√(1/4) =

- √1 / √4 =

- 1 / 2

The flight of the ball is a parabola.

If we take the first derivative of the function, we can find the time when the ball reaches a maximum height

f'(x) = -10x +40      And setting this to 0, we have

-10x + 40 = 0         x = 4 sec

And it will take 4 more seconds to land, so it was in the air for 8 secs.

Verify this: At 0 secs. the height was 0, and at 8 secs. the height is, -5*8^2 + 40*8 = 0