Question

Use the graph to determine a. The function's domain; b. The functions range; c. The x-intercepts, if any; d. The y-intercept, if any; and e. The missing function values, indicated by question marks, below. F(-2)=? F(2)=?

Answer 1

Answer:

(a) Domain is $(-\infty,+\infty)$ (b) Range is $(-\infty,+1]$ (c) x intercepts  -1 and +1 (d) y-intercept is +1

(e)

$F(-2)=-2\\F(2)=-2$

Step-by-step explanation:

(a) The function's domain:

From the given graph, it is clear that the value of function exist in the interval from $(-\infty,+\infty)$.

Therefore, the domain of the function is $(-\infty,+\infty)$.

(b) The function's range:

Clearly from the graph that is given in the question, the highest value of y coordinate is +1 and its lowest value is $-\infty$.

Hence, the range of the function is $(-\infty,+1]$

(c) The x-intercepts:

The curve intersects the x-axis at -1 and +1, therefore the intercepts of the curve is -1 and +1.

(d) The y-intercept:

The curve intersect the y-axis at y=1, therefore y-intercept is +1.

(e) Missing Values:

Clearly from graph:

$F(-2)=-2\\F(2)=-2$

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Answer 2

Utilizing the graph, it is found that:

a) The domain of the function is all real values.

b) The range is given by: $(-\infty, 1]$

c) The x-intercepts are $x = \pm 1$

d) The y-intercept is $y = 1$

e) $F(-2) = F(2) = -3$

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Item a:

• The domain of a function is given by all possible values of the input.
• On the graph, it is the values assumed by the x-axis, the horizontal axis.
• On the graph, the function is defined for all values on the horizontal axis, thus, the domain is all real values.

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Item b:

• The range of a function is given by all possible values of the output.
• On the graph, it is the values assumed by the y-axis, the vertical axis.
• On the graph, the function assumes values less than 1, thus, the range is: $(-\infty, 1]$

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Item c:

• A x-intercept is given by the value of x when y = 0.
• On the graph, it is the values of x when it crosses the horizontal axis.
• Looking at the graph, it is found that these values are $x = \pm 1$

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Item d:

• A y-intercept is given by the value of y when x = 0.
• On the graph, it is the values of y when it crosses the vertical axis.

• Looking at the graph, it is found that this value is $y = 1$

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Item e:

• $F(-2) = F(2)$ are given by the values of y when x = -2 and 2, that is, the vertical axis when the horizontal axis is -2 and 2.
• Looking at the graph, we find that in both cases, y = -3, thus $F(-2) = F(2) = 3$

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