Graph the function f(x) = x3 – 4x – 1. Which are approximate solutions for x when f(x) = 0? Check all that apply. –2.11 –1.86 –0
.25 0.25 1.86 2.11
2 answers:
Answer:
-1.86, -0.25 and 2.11
Step-by-step explanation:
In the figure attached a plot of the function is shown. The solutions for x when f(x) = 0 are also given.
To check if a given x is solution of f(x) = 0, just replace x value in the formula and verify if the function is or not equal to zero.
x f(x)
-2.11 (-2.11)^3 - 4(-2.11) -1 = -1.95
-1.86 (-1.86)^3 - 4(-1.86) -1 ≅ 0
-0.25 (-0.25)^3 - 4(-0.25) -1 ≅ 0
0.25 (0.25)^3 - 4(0.25) -1 = -1.98
1.86 (1.86)^3 - 4(1.86) -1 = -2
2.11 (2.11)^3 - 4(2.11) -1 ≅ 0
Therefore, the solutions are: -1.86, -0.25 and 2.11
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Answer:
- positive
- negative
Step-by-step explanation:
The thrust of the question is to make sure you understand that increasing the y-coordinate of a point will move the point upward, and decreasing it will move the point downward.
That is adding a positive value "k" to x^2 will move the point (x, x^2) to the point (x, x^2+k), which will be above the previous point by k units.
If k is subtracted, instead of added, then the point will be moved downward.
The blanks are supposed to be filled with <u> positive </u>, and <u> negative </u>.
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<em>Comment on the question</em>
The wording of the statement you're completing is a bit odd. If k is negative (-2, for example), this statement is saying the graph is translated down -2 units. It is not. It is translated down |-2| = 2 units. The direction of translation depends on the sign of k. The amount of translation depends on the magnitude of k.
If you thoroughly understand (x, y) coordinates and how they are plotted on a graph, it should be no mystery that changing the y-coordinate will change the vertical position of the graph.
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And we're done.
Thanks for watching buddy good luck.
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