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Answer:</h3>
y = (x +2)(x -1)(x -3) . . . . or . . . . y = x³ -2x² -5x +6
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Step-by-step explanation:</h3>
The graph shows y=0 at x=-2, x=1, and x=3. These are called the "zeros" or "roots" of the function, because the value of the function is zero there.
When "a" is a zero of a polynomial function, (x -a) is a factor. This means the factors of the graphed function are (x -(-2)), (x -1) and (x -3). The function can be written as the product of these factors:
... y = (x +2)(x -1)(x -3) . . . . . the equation represented by the graph
Or, the product can be multiplied out
... y = (x +2)(x² -4x +3)
... y = x³ -2x² -5x +6 . . . . . the equation represented by the graph
Answer:
you will 180 pints of the 100% juice and 540 pints of the 40% juice
Step-by-step explanation:
in order to get a 55% mixture you must combine 1 100% and 3 40%
100 + 120 = 220 then divide by 4 and you get 55% juice
now you must use 180 100% juice and 540 40 % juice to make 180 pints
you need this because the ratio is 3 40% to every 1 100%
Answer:
a. A(x) = (1/2)x(9 -x^2)
b. x > 0 . . . or . . . 0 < x < 3 (see below)
c. A(2) = 5
d. x = √3; A(√3) = 3√3
Step-by-step explanation:
a. The area is computed in the usual way, as half the product of the base and height of the triangle. Here, the base is x, and the height is y, so the area is ...
A(x) = (1/2)(x)(y)
A(x) = (1/2)(x)(9-x^2)
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b. The problem statement defines two of the triangle vertices only for x > 0. However, we note that for x > 3, the y-coordinate of one of the vertices is negative. Straightforward application of the area formula in Part A will result in negative areas for x > 3, so a reasonable domain might be (0, 3).
On the other hand, the geometrical concept of a line segment and of a triangle does not admit negative line lengths. Hence the area for a triangle with its vertex below the x-axis (green in the figure) will also be considered to be positive. In that event, the domain of A(x) = (1/2)(x)|9 -x^2| will be (0, ∞).
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c. A(2) = (1/2)(2)(9 -2^2) = 5
The area is 5 when x=2.
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d. On the interval (0, 3), the value of x that maximizes area is x=√3. If we consider the domain to be all positive real numbers, then there is no maximum area (blue dashed curve on the graph).