Ryan deposited $100 into his bank account. Later he withdrew $75 from the account. Write an expression that represents the chang
1 answer:
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Answer:
Look to the attached graph
Step-by-step explanation:
* Lets revise how to graph the quadratic function
- Find the vertex of it
- Find the y-intercept
- Find the x-intercept
∵ f(x) = (x - 5)² + 1 ⇒ the completing square form
- The completing square form for any quadratic is
( x - h)² + k, where h and k are the coordinate of the vertex point
* Lets compare the two forms
∵ (x - h)² + k = (x - 5)² + 1
∴ h = 5 and k = 1
∴ The vertex of the parabola is (5 , 1)
- To find the x-intercept put f(x) = 0
∵ (x - 5)² + 1 = 0 ⇒ subtract 1 from both sides
∴ (x - 5)² = -1 ⇒ the square can not give -ve number
∴ The parabola does not intersect the x-axis
- To find the y-intercept put x = 0
∵ f(0) = (0 - 5)² + 1 = 25 + 1 = 26
∴ The parabola intersects the y-axis at point (0 , 26)
- The parabola is opened upward because the coefficient of x² is +ve
* Now lets graph it
- Look to the attached graph
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).
