The equation of a curve is y=16−ax−x2y=16-ax-x^{2}, where aa is positive constant.

  1. (a)

    Given that yy can be expressed in the form 25−(b+x)225-(b+x)^{2}, where bb is a positive constant, find the values of aa and bb.

  2. (b)

    State the maximum value of yy and the corresponding value of xx.

  3. (c)

    Find the range of values of xx when yy is positive.

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Worked solution

11(a)

1. Expand the given completed square form

25−(b+x)2=25−(b2+2bx+x2)25 - (b+x)^2 = 25 - (b^2 + 2bx + x^2)
=(25−b2)−2bx−x2= (25 - b^2) - 2bx - x^2

2. Compare the constant term to find $b$

Comparing the constant term with 16−ax−x216 - ax - x^2:
25−b2=1625 - b^2 = 16
b2=9b^2 = 9
Since bb is a positive constant, b=3b = 3.

3. Compare the coefficient of $x$ to find $a$

Comparing the coefficient of xx:
−2b=−a-2b = -a
a=2(3)=6a = 2(3) = 6

Answer

a=6a = 6, b=3b = 3

11(b)

1. Find the value of $b$ from part (a) to complete the expression for $y$

Expanding 25−(b+x)225 - (b+x)^2 gives 25−b2−2bx−x225 - b^2 - 2bx - x^2.
Comparing the constant term with y=16−ax−x2y = 16 - ax - x^2:
25−b2=1625 - b^2 = 16
b2=9b^2 = 9
Since bb is positive, b=3b = 3.
Thus, y=25−(x+3)2y = 25 - (x+3)^2.

2. State the maximum value of $y$ and the corresponding value of $x$

From y=25−(x+3)2y = 25 - (x+3)^2, the maximum value occurs when the squared term is zero, i.e., (x+3)2=0(x+3)^2 = 0.
Maximum value of y=25y = 25.
This occurs when x+3=0  ⟹  x=−3x + 3 = 0 \implies x = -3.

Answer

Maximum value of y=25y = 25, corresponding value of x=−3x = -3

11(c)

1. Set up the inequality for $y > 0$ using the expression from part (a)

From part (a), y=25−(x+3)2y = 25 - (x+3)^2.
For yy to be positive, 25−(x+3)2>025 - (x+3)^2 > 0.

2. Solve the quadratic inequality

(x+3)2<25(x+3)^2 < 25
−5<x+3<5-5 < x+3 < 5

3. Find the final range of values for $x$

−5−3<x<5−3-5 - 3 < x < 5 - 3
−8<x<2-8 < x < 2

Answer

−8<x<2-8 < x < 2

AI analysis — may contain mistakes.

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