The curve y = 11x - 24 - x^2 cuts the x-axis at points A and B, and PN is the greatest positive value of the y coordinate. Show that 2 PN • AB equals three times the area bounded by that portion of the curve which lies in the first quadrant.
Kinda confused on how to do this
So you want me to do it three times?
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To solve this problem, there are several steps involved. Let's break it down:
Step 1: Find the x-coordinate of points A and B.
To find the x-coordinates of points A and B, we need to set the equation of the curve, y = 11x - 24 - x^2, equal to zero since the curve cuts the x-axis at those points. So, we solve the equation:
11x - 24 - x^2 = 0
This equation is a quadratic equation in the form ax^2 + bx + c = 0. To solve it, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = -1, b = 11, and c = -24. Plugging these values into the quadratic formula, we get:
x = (-(11) ± √((11)^2 - 4(-1)(-24))) / (2(-1))
Simplifying further:
x = (-11 ± √(121 - 96)) / (-2)
x = (-11 ± √25) / (-2)
So, there are two possible x-values for points A and B: x = 3 and x = -8.
Step 2: Calculate the y-coordinate of the maximum point (P).
To find the y-coordinate of the maximum point on the curve (P), we need to find the vertex of the parabola. The vertex of a quadratic equation in the form y = ax^2 + bx + c is given by the formula:
x_vertex = -b / (2a)
y_vertex = f(x_vertex)
For our equation, a = -1 and b = 11. Plugging these values into the formulas, we have:
x_vertex = -(11) / (2(-1))
x_vertex = 11/2
To find the y-coordinate of the vertex, substitute the x_vertex back into the original equation:
y_vertex = 11(11/2) - 24 - (11/2)^2
y_vertex = 121/2 - 24 - 121/4
y_vertex = 242/4 - 96/4 - 121/4
y_vertex = 25/4
So, the y-coordinate of point P is 25/4.
Step 3: Calculate the length of AB.
The length of AB can be calculated using the x-coordinates of points A and B, which we found earlier. The length of a line segment is given by the formula:
AB = |x_B - x_A|
Since A has the x-coordinate 3 and B has the x-coordinate -8, we have:
AB = |-8 - 3| = 11
Step 4: Calculate the area bounded by the portion of the curve in the first quadrant.
To find the area bounded by the portion of the curve in the first quadrant, we need to integrate the equation of the curve from x = 0 to the x-coordinate of point A (x = 3). Recall that the area under a curve can be found using integration.
The equation of the curve is y = 11x - 24 - x^2, so the area can be calculated as follows:
Area = ∫[0 to 3] (11x - 24 - x^2) dx
Evaluating the integral gives:
Area = [11x^2/2 - 24x - x^3/3] from 0 to 3
Area = [11(3)^2/2 - 24(3) - (3)^3/3] - [11(0)^2/2 - 24(0) - (0)^3/3]
Area = [99/2 - 72 - 9] - [0 - 0 - 0]
Area = 99/2 - 72 - 9
Area = 99/2 - 153/2
Area = -54/2
Area = -27
Since we are interested in the positive area, we take the absolute value:
Area = | -27 | = 27
Step 5: Calculate 2PN • AB.
Now, we can calculate 2PN • AB using the values we obtained earlier.
2PN • AB = 2(25/4) • 11
2PN • AB = 275/2
Step 6: Check if 2PN • AB equals three times the bounded area.
Finally, we check if 2PN • AB is equal to three times the bounded area:
Is 2PN • AB = 3 • Area?
Is 275/2 = 3 • 27?
Is 275/2 = 81/2?
No, 275/2 ≠ 81/2.
Therefore, the statement "2PN • AB equals three times the area bounded by that portion of the curve which lies in the first quadrant" is not true based on the calculations we just did.
I hope that helps! Let me know if you have any other questions.