>>667577q(4,-20) lies on the curve. Therefore:

-20 = 2*a*sqrt(4+b)

-20/2 = a*sqrt(4+b)

-10 = a*sqrt(4+b)

(-10)^2 = (a*sqrt(4+b))^2

100 = a^2*(4+b)

First equation: 100 = a^2*(4+b)

now we know the tangent slope of y at Q(4,-20) is -3, or in other words y' = -3 for Q(4-20).

Assuming you know how to differentiate a function, we have:

y' = a/sqrt(x+b)

Using x = 4 and y' = -3 (the slope given, which is at x = 4), we have:

-3 = a/sqrt(4+b)

-3sqrt(4+b)=a

(-3sqrt(4+b))^2=(a)^2

9(4+b) = a^2

a^2 = 9(4+b)

a^2 = 36+9b

Second equation: a^2 = 36+9b

Now we backtrack, using the definition of a given by the second equation, and plug it in the first equation 100 = a^2*(4+b), where a^2 is 36+9b (our second equation):

100 = a^2*(4+b)

100 = (36+9b)*(4+b)

100 = 144+36b+36b+9b^2

9b^2+72b+144 = 100

9b^2+72b+144-100 = 0

9b^2+72b+44 = 0

Solving the above quadratic formula, we have:

b1 =-22/3

b2 = -2/3

Now, although we have 2 solutions, we have to figure out which one is the correct one. This is done by looking at the original function and the point given.

y = 2asqrt(x+b)

Now we know that there is no real solution to the square root of a negative number, therefore our solution for b cannot yield a negative square root. We know that our point Q(4,-20) is on the curve of our function. Therefore, we try using both b found:

Replacing b = -22/3, we have:

y = 2asqrt(x-22/3)

Now at x = 4, our function must yield a real value. But sqrt(4-22/3) = sqrt(-3.3333) = no solution.

Therefore b cannot be -22/3.

We do the same thing with b = -2/3.

Replacing b = -2/3, we have:

y = 2asqrt(x-2/3)

Now at x = 4, our function must yield a real value. sqrt(4-2/3) = sqrt(3.3333) = real number.

We now know that b = -2/3.

Now we must find a. We once again go back to our original function, plugging in for x, y and b, and we have:

y = 2*a*sqrt(x-b)

-20 = 2*a*sqrt(4-2/3)

-20/2 = a*sqrt(10/3)

-10 = a*sqrt(10/3)

-10/sqrt(10/3) = a

a = -5.477225575...

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