The perimeter of a rectangle is 40 feet. If the length is 15 feet, 6 inches ; what is the width
1 answer:
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(a) See the first attachment for a graph. This graphing calculator displays roots to 3 decimal places. (The third attachment shows a different graphing calculator and 10 significant digits.)
(b) In the table of the first attachment, the column headed by g(x) gives iterations of Newton's Method. (For Newton's method, it is convenient to let the calculator's derivative function compute the derivative f'(x) of the function f(x). We have defined g(x) = x - f(x)/f'(x).) The result of the 3rd iteration is ...
... x ≈ 3.0473167
(c) The function h(x₁, x₂) computes iterations using the secant method. The results for three iterations of that method are shown below the table in the attachment. The result of the 3rd iteration is ...
... x ≈ 3.2291234
(d) The function h(x, x+0.01) computes the modified secant method as required by the problem statement. The result of the 3rd iteration is ...
... x ≈ 3.0477377
(e) Using <em>Mathematica</em>, the roots are found to be as shown in the second attachment. The highest root is about ...
... x ≈ 3.0466805180
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<em>Comment on these methods</em>
Newton's method can have convergence problems if the starting point is not sufficiently close to the root. A graphing calculator that gives a 3-digit approximation (or better) can help avoid this issue. For the calculator used here, the output of "g(x)" is computed even as the input is typed, so one can simply copy the function output to the input to get a 12-significant digit approximation of the root as fast as you can type it.
The "modified" secant method is a variation of the secant method that does not require two values of the function to start with. Instead, it uses a value of x that is "close" to the one given. For our purpose here, we can use the same h(x1, x2) for both methods, with a different x2 for the modified method.
We have defined h(x1, x2) = x1 - f(x1)(f(x1)-f(x2))/(x1 -x2).
Answer:
y = -3(x - 4)² - 2
Step-by-step explanation:
Given the vertex, (4, -2), and the point (2, -14):
We can use the vertex form of the quadratic equation:
y = a(x - h)² + k
Where:
(h, k) = vertex
a = determines whether the graph opens up or down, and it also makes the parent function <u>wider</u> or <u>narrower</u>.
- <u>positive</u> value of a = opens <u><em>upward</em></u>
- <u>negative</u> value of a = opens <u><em>downward</em></u>
- a is between 0 and 1, (0 < a < 1) the graph is <u><em>wider</em></u> than the parent function.
- a > 1, the graph is <u><em>narrower</em></u> than the parent function.
<em>h </em>=<em> </em>determines how far left or right the parent function is translated.
- h = positive, the function is translated <em>h</em> units to the right.
- h = negative, the function is translated |<em>h</em>| units to the left.
<em>k</em> determines how far up or down the parent function is translated.
- k = positive: translate <em>k</em> units <u><em>up</em></u>.
- k = negative, translate <em>k</em> units <u><em>down</em></u>.
Now that I've set up the definitions for each variable of the vertex form, we can determine the quadratic equation using the given vertex and the point:
vertex (h, k): (4, -2)
point (x, y): (2, -14)
Substitute these values into the vertex form to solve for a:
y = a(x - h)² + k
-14 = a(2 - 4)² -2
-14 = a (-2)² -2
-14 = a4 + -2
Add to to both sides:
-14 + 2 = a4 + -2 + 2
-12 = 4a
Divide both sides by 4 to solve for a:
-12/4 = 4a/4
-3 = a
Therefore, the quadratic equation inI vertex form is:
y = -3(x - 4)² - 2
The parabola is downward-facing, and is vertically compressed by a factor of -3. The graph is also horizontally translated 4 units to the right, and vertically translated 2 units down.
Attached is a screenshot of the graph where it shows the vertex and the given point, using the vertex form that I came up with.
Please mark my answers as the Brainliest, if you find this helpful :)
