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(1.) Write these inequalities in interval notation.

$ (a.)\;\;-3 \lt x \lt 5 \\[3ex] (b.)\;\;-3 \le x \lt 5 \\[3ex] (c.)\;\;-3 \lt x \le 5 \\[3ex] (d.)\;\;-3 \le x \le 5 \\[3ex] (e.)\;\; x \lt 5 \\[3ex] (f.)\;\; x \gt 5 \\[3ex] (g.)\;\; x \le 5 \\[3ex] (h.)\;\; 5 \ge x \\[3ex] (i)\;\; 5 \gt x \\[3ex] (j.)\;\; 5 \lt x \\[3ex] (k.)\;\; 5 \le x \\[3ex] $

$ (a.) \\[3ex] -3 \lt x \lt 5 \\[3ex] (-3, 5) \\[3ex] (b.) \\[3ex] -3 \le x \lt 5 \\[3ex] [-3, 5) \\[3ex] (c.) \\[3ex] -3 \lt x \le 5 \\[3ex] (-3, 5] \\[3ex] (d.) \\[3ex] -3 \le x \le 5 \\[3ex] [-3, 5] \\[3ex] (e.) \\[3ex] x \lt 5 \\[3ex] (-\infty, 5) \\[3ex] (f.) \\[3ex] x \gt 5 \\[3ex] (5, \infty) \\[3ex] (g.) \\[3ex] x \le 5 \\[3ex] (-\infty, 5] \\[3ex] (h.) \\[3ex] 5 \ge x \;\;means\;\;that\;\; x \le 5 \\[3ex] (-\infty, 5] \\[3ex] (i.) \\[3ex] 5 \gt x \implies x \lt 5 \\[3ex] (-\infty, 5) \\[3ex] (j.) \\[3ex] 5 \lt x \implies x \gt 5 \\[3ex] (5, \infty) \\[3ex] (k.) \\[3ex] 5 \le x \implies x \ge 5 \\[3ex] [5, \infty) $

$ (a.)\;\;-3 \lt x \lt 5 \\[3ex] (b.)\;\;-3 \le x \lt 5 \\[3ex] (c.)\;\;-3 \lt x \le 5 \\[3ex] (d.)\;\;-3 \le x \le 5 \\[3ex] (e.)\;\; x \lt 5 \\[3ex] (f.)\;\; x \gt 5 \\[3ex] (g.)\;\; x \le 5 \\[3ex] (h.)\;\; 5 \ge x \\[3ex] (i)\;\; 5 \gt x \\[3ex] (j.)\;\; 5 \lt x \\[3ex] (k.)\;\; 5 \le x \\[3ex] $

$ (a.) \\[3ex] -3 \lt x \lt 5 \\[3ex] (-3, 5) \\[3ex] (b.) \\[3ex] -3 \le x \lt 5 \\[3ex] [-3, 5) \\[3ex] (c.) \\[3ex] -3 \lt x \le 5 \\[3ex] (-3, 5] \\[3ex] (d.) \\[3ex] -3 \le x \le 5 \\[3ex] [-3, 5] \\[3ex] (e.) \\[3ex] x \lt 5 \\[3ex] (-\infty, 5) \\[3ex] (f.) \\[3ex] x \gt 5 \\[3ex] (5, \infty) \\[3ex] (g.) \\[3ex] x \le 5 \\[3ex] (-\infty, 5] \\[3ex] (h.) \\[3ex] 5 \ge x \;\;means\;\;that\;\; x \le 5 \\[3ex] (-\infty, 5] \\[3ex] (i.) \\[3ex] 5 \gt x \implies x \lt 5 \\[3ex] (-\infty, 5) \\[3ex] (j.) \\[3ex] 5 \lt x \implies x \gt 5 \\[3ex] (5, \infty) \\[3ex] (k.) \\[3ex] 5 \le x \implies x \ge 5 \\[3ex] [5, \infty) $

(2.) Write these intervals in set notation:

$ (a.)\;\; (7, 9) \\[3ex] $

$ (a.) \\[3ex] (7,9) = \{x| 7 \lt x \lt 9\} \\[3ex] $

$ (a.)\;\; (7, 9) \\[3ex] $

$ (a.) \\[3ex] (7,9) = \{x| 7 \lt x \lt 9\} \\[3ex] $

(3.) (a.) Plot the points (3, −1), (6, −9), (9, −6), (3, −5) in the Cartesian plane.

Choose the correct graph.

(b.) Does the relation represent a function? Why?

**A.** Yes, because there are ordered pairs with the same first element and different second elements.

**B.** Yes, because there are no ordered pairs with the same first element and different second elements.

**C.** No, because there are no ordered pairs with the same first element and different second elements.

**D.** No, because there are ordered pairs with the same first element and different second elements.

(a.) Scale: 1 cm represents 2 units (Not drawn to scale)

The correct graph representing the points is Option**D.**:

(b.) The relation is__not__ a function because there is an input, 3 that has two different outputs, −1 and −5

**D.** No, because there are ordered pairs with the same first element and different second elements.

Choose the correct graph.

(b.) Does the relation represent a function? Why?

(a.) Scale: 1 cm represents 2 units (Not drawn to scale)

The correct graph representing the points is Option

(b.) The relation is

(4.)

(5.)

(6.) Which of the following intervals is required to guarantee a continuous function will have both an absolute maximum and an absolute minimum?

$ A.\;\; (a, b) \\[3ex] B.\;\; [a, b] \\[3ex] C.\;\; [a, b) \\[3ex] D.\;\; (a, b] \\[3ex] $

The correct option is**B.**

If*f* is a continuous function, whose domain is a closed interval [a,b] , then *f* has an absolute maximum and an absolute minimum on [a,b].

$ A.\;\; (a, b) \\[3ex] B.\;\; [a, b] \\[3ex] C.\;\; [a, b) \\[3ex] D.\;\; (a, b] \\[3ex] $

The correct option is

If

(7.)

(8.)

(9.) Determine the intercepts of these equations:

$ (a.)\;\; y = 6x^2 - 6 \\[3ex] (b.)\;\; 9x^2 + 4y^2 = 36 \\[3ex] $

To find: (1.) the*x*-intercept, set *y* = 0 __and__ solve for *x*

(*x*-values, 0) gives the *x*-intercept

(2.) the*y*-intercept, set *x* = 0 __and__ solve for *y*

(0,*y*-values) gives the *y*-intercept

$ (a.) \\[3ex] y = 8x^2 - 8 \\[5ex] \underline{y-intercept} \\[3ex] y = 8(0)^2 - 8 \\[3ex] y = 0 - 8 \\[3ex] y = -8 \\[3ex] y-intercept = (0, -8) \\[3ex] \underline{x-intercept} \\[3ex] 0 = 8x^2 - 8 \\[3ex] 8x^2 - 8 = 0 \\[3ex] 8(x^2 - 1) = 0 \\[3ex] x^2 - 1 = 0 \\[3ex] x^2 = 0 + 1 \\[3ex] x^2 = 1 \\[3ex] x = \pm\sqrt{1} \\[3ex] x = \pm 1 \\[3ex] x-intercepts = (-1, 0) \;\;\;and\;\;\; (1, 0) \\[5ex] (b.) \\[3ex] 9x^2 + 4y^2 = 36 \\[3ex] \underline{y-intercept} \\[3ex] 9(0)^2 + 4y^2 = 36 \\[3ex] 0 + 4y^2 = 36 \\[3ex] 4y^2 = 36 \\[3ex] y^2 = \dfrac{36}{4} \\[5ex] y^2 = 9 \\[3ex] y = \pm \sqrt{9} \\[3ex] y = \pm 3 \\[3ex] y-intercepts = (0, -3) \;\;\;and\;\;\; (0, 3) \\[3ex] \underline{x-intercept} \\[3ex] 9x^2 + 4(0)^2 = 36 \\[3ex] 9x^2 + 0 = 36 \\[3ex] 9x^2 = 36 \\[3ex] x^2 = \dfrac{36}{9} \\[5ex] x^2 = 4 \\[3ex] x = \pm \sqrt{4} \\[3ex] x-intercepts = (-2, 0) \;\;\;and\;\;\; (2, 0) \\[5ex] $

$ (a.)\;\; y = 6x^2 - 6 \\[3ex] (b.)\;\; 9x^2 + 4y^2 = 36 \\[3ex] $

To find: (1.) the

(

(2.) the

(0,

$ (a.) \\[3ex] y = 8x^2 - 8 \\[5ex] \underline{y-intercept} \\[3ex] y = 8(0)^2 - 8 \\[3ex] y = 0 - 8 \\[3ex] y = -8 \\[3ex] y-intercept = (0, -8) \\[3ex] \underline{x-intercept} \\[3ex] 0 = 8x^2 - 8 \\[3ex] 8x^2 - 8 = 0 \\[3ex] 8(x^2 - 1) = 0 \\[3ex] x^2 - 1 = 0 \\[3ex] x^2 = 0 + 1 \\[3ex] x^2 = 1 \\[3ex] x = \pm\sqrt{1} \\[3ex] x = \pm 1 \\[3ex] x-intercepts = (-1, 0) \;\;\;and\;\;\; (1, 0) \\[5ex] (b.) \\[3ex] 9x^2 + 4y^2 = 36 \\[3ex] \underline{y-intercept} \\[3ex] 9(0)^2 + 4y^2 = 36 \\[3ex] 0 + 4y^2 = 36 \\[3ex] 4y^2 = 36 \\[3ex] y^2 = \dfrac{36}{4} \\[5ex] y^2 = 9 \\[3ex] y = \pm \sqrt{9} \\[3ex] y = \pm 3 \\[3ex] y-intercepts = (0, -3) \;\;\;and\;\;\; (0, 3) \\[3ex] \underline{x-intercept} \\[3ex] 9x^2 + 4(0)^2 = 36 \\[3ex] 9x^2 + 0 = 36 \\[3ex] 9x^2 = 36 \\[3ex] x^2 = \dfrac{36}{9} \\[5ex] x^2 = 4 \\[3ex] x = \pm \sqrt{4} \\[3ex] x-intercepts = (-2, 0) \;\;\;and\;\;\; (2, 0) \\[5ex] $

(10.)

(11.) Determine whether the graph is that of a function by using the vertical-line test.

If it is a function, use the graph to find

(a.) its domain and range.

(b.) the intercepts, if any.

(c.) any symmetry with respect to the x-axis, y-axis, or the origin.

The graph failed the Vertical Line Test because the vertical line: $x = 6$ intersects the graph at two points $(6, 2)$ and $(6, -2)$

The same input, 6 has two different outputs: 2 and −2

It is__not__ a function.

If it is a function, use the graph to find

(a.) its domain and range.

(b.) the intercepts, if any.

(c.) any symmetry with respect to the x-axis, y-axis, or the origin.

The graph failed the Vertical Line Test because the vertical line: $x = 6$ intersects the graph at two points $(6, 2)$ and $(6, -2)$

The same input, 6 has two different outputs: 2 and −2

It is

(12.)

(13.)

(14.)

(15.) Determine whether the graph is that of a function by using the vertical-line test.

If it is a function, use the graph to find

(a.) the domain and range.

(b.) the intercepts, if any.

(c.) any symmetry with respect to the*x*-axis, *y*-axis, or the origin.

The graph passed the Vertical Line Test because the vertical line: $x = \dfrac{\pi}{2}$ intersects the graph at only one point $\left(\dfrac{\pi}{2}, -1\right)$

It is a function.

It is a cubic function.

It is a polynomial function.

$ (a.) \\[3ex] \underline{Domain} \\[3ex] Minimum\;\;x-value = -\pi \\[3ex] Maximum x-value = \pi \\[3ex] D = [-\pi, \pi] \\[3ex] \underline{Range} \\[3ex] Minimum\;\;y-value = -1 \\[3ex] Maximum\;\;y-value = 1 \\[3ex] R = [-1, 1] \\[3ex] (b.) \\[3ex] \underline{x-intercepts} \\[3ex] x-intercepts = (-\pi, 0) \;\;\;and\;\;\; (\pi, 0) \\[3ex] \underline{y-intercept} \\[3ex] y-intercept = (0, 0) \\[3ex] $ (c.) Let us check each symmetry.

**Symmetric about the ***x*-axis

For each $(x, y)$ on the graph; there is also $(x, -y)$ on the same graph

Same $x$, Opposite $y$

We have $\left(\dfrac{\pi}{2}, -1\right)$ but we do not have $\left(\dfrac{\pi}{2}, 1\right)$ on the graph.

The graph is__not__ symmetric about the *x*-axis

**Symmetric about the ***y*-axis

For each $(x, y)$ on the graph; there is also $(−x, y)$ on the same graph

Same $y$, Opposite $x$

We have $\left(\dfrac{\pi}{2}, -1\right)$ but we do not have $\left(-\dfrac{\pi}{2}, -1\right)$ on the graph.

The graph is__not__ symmetric about the *y*-axis

**Symmetric about the origin**

For each $(x, y)$ on the graph; there is also $(−x, −y)$ on the same graph

Opposite $x$, Opposite $y$

We have $\left(\dfrac{\pi}{2}, -1\right)$__and__ we also have $\left(-\dfrac{\pi}{2}, 1\right)$ on the graph.

The graph is symmetric about the origin.

If it is a function, use the graph to find

(a.) the domain and range.

(b.) the intercepts, if any.

(c.) any symmetry with respect to the

The graph passed the Vertical Line Test because the vertical line: $x = \dfrac{\pi}{2}$ intersects the graph at only one point $\left(\dfrac{\pi}{2}, -1\right)$

It is a function.

It is a cubic function.

It is a polynomial function.

$ (a.) \\[3ex] \underline{Domain} \\[3ex] Minimum\;\;x-value = -\pi \\[3ex] Maximum x-value = \pi \\[3ex] D = [-\pi, \pi] \\[3ex] \underline{Range} \\[3ex] Minimum\;\;y-value = -1 \\[3ex] Maximum\;\;y-value = 1 \\[3ex] R = [-1, 1] \\[3ex] (b.) \\[3ex] \underline{x-intercepts} \\[3ex] x-intercepts = (-\pi, 0) \;\;\;and\;\;\; (\pi, 0) \\[3ex] \underline{y-intercept} \\[3ex] y-intercept = (0, 0) \\[3ex] $ (c.) Let us check each symmetry.

For each $(x, y)$ on the graph; there is also $(x, -y)$ on the same graph

Same $x$, Opposite $y$

We have $\left(\dfrac{\pi}{2}, -1\right)$ but we do not have $\left(\dfrac{\pi}{2}, 1\right)$ on the graph.

The graph is

For each $(x, y)$ on the graph; there is also $(−x, y)$ on the same graph

Same $y$, Opposite $x$

We have $\left(\dfrac{\pi}{2}, -1\right)$ but we do not have $\left(-\dfrac{\pi}{2}, -1\right)$ on the graph.

The graph is

For each $(x, y)$ on the graph; there is also $(−x, −y)$ on the same graph

Opposite $x$, Opposite $y$

We have $\left(\dfrac{\pi}{2}, -1\right)$

The graph is symmetric about the origin.

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(23.) Determine whether the graph is that of a function by using the vertical-line test.

If it is a function, use the graph to find

(a.) its domain and range.

(b.) the intercepts, if any.

(c.) any symmetry with respect to the x-axis, y-axis, or the origin.

The graph failed the Vertical Line Test because the vertical line: $x = -1$ intersects the graph at two points $(-1, 1)$ and $(-1, -1)$

The same input, −1 has two different outputs: 1 and −1

It is__not__ a function.

If it is a function, use the graph to find

(a.) its domain and range.

(b.) the intercepts, if any.

(c.) any symmetry with respect to the x-axis, y-axis, or the origin.

The graph failed the Vertical Line Test because the vertical line: $x = -1$ intersects the graph at two points $(-1, 1)$ and $(-1, -1)$

The same input, −1 has two different outputs: 1 and −1

It is

(24.)

(25.)

(26.)

(27.) Determine whether the graph is that of a function by using the vertical-line test.

If it is a function, use the graph to find

(a.) the domain and range (Assume that the curve approaches but never intersects the*y*-axis).

(b.) the intercepts, if any.

(c.) any symmetry with respect to the*x*-axis, *y*-axis, or the origin.

The graph passed the Vertical Line Test because the vertical line: $x = 1$ intersects the graph at only one point $(1, 0)$

It is a function.

$ (a.) \\[3ex] \underline{Domain} \\[3ex] Minimum\;\;x-value = 0(excluded) \\[3ex] Maximum x-value = 2(excluded) \\[3ex] D = (0, 2) \\[3ex] \underline{Range} \\[3ex] Minimum\;\;y-value = -\infty(excluded) \\[3ex] Maximum\;\;y-value = 2 \\[3ex] R = (-\infty, 2) \\[3ex] (b.) \\[3ex] \underline{x-intercept} \\[3ex] x-intercept = (1, 0) \\[3ex] \underline{y-intercept} \\[3ex] y-intercept = None \\[3ex] $ (c.) Let us check each symmetry.

**Symmetric about the ***x*-axis

For each $(x, y)$ on the graph; there is also $(x, -y)$ on the same graph

Same $x$, Opposite $y$

We have $(2, 2)$ but we do not have $(2, -2)$ on the graph.

The graph is__not__ symmetric about the *x*-axis

**Symmetric about the ***y*-axis

For each $(x, y)$ on the graph; there is also $(−x, y)$ on the same graph

Same $y$, Opposite $x$

We have $(1, 0)$ but we do not have $(-1, 0)$ on the graph.

The graph is__not__ symmetric about the *y*-axis

**Symmetric about the origin**

For each $(x, y)$ on the graph; there is also $(−x, −y)$ on the same graph

Opposite $x$, Opposite $y$

We have $(2, 2)$ but we do not have $(-2, -2)$ on the graph.

The graph is__not__ symmetric about the origin.

Therefore, the graph has no symmetry.

If it is a function, use the graph to find

(a.) the domain and range (Assume that the curve approaches but never intersects the

(b.) the intercepts, if any.

(c.) any symmetry with respect to the

The graph passed the Vertical Line Test because the vertical line: $x = 1$ intersects the graph at only one point $(1, 0)$

It is a function.

$ (a.) \\[3ex] \underline{Domain} \\[3ex] Minimum\;\;x-value = 0(excluded) \\[3ex] Maximum x-value = 2(excluded) \\[3ex] D = (0, 2) \\[3ex] \underline{Range} \\[3ex] Minimum\;\;y-value = -\infty(excluded) \\[3ex] Maximum\;\;y-value = 2 \\[3ex] R = (-\infty, 2) \\[3ex] (b.) \\[3ex] \underline{x-intercept} \\[3ex] x-intercept = (1, 0) \\[3ex] \underline{y-intercept} \\[3ex] y-intercept = None \\[3ex] $ (c.) Let us check each symmetry.

For each $(x, y)$ on the graph; there is also $(x, -y)$ on the same graph

Same $x$, Opposite $y$

We have $(2, 2)$ but we do not have $(2, -2)$ on the graph.

The graph is

For each $(x, y)$ on the graph; there is also $(−x, y)$ on the same graph

Same $y$, Opposite $x$

We have $(1, 0)$ but we do not have $(-1, 0)$ on the graph.

The graph is

For each $(x, y)$ on the graph; there is also $(−x, −y)$ on the same graph

Opposite $x$, Opposite $y$

We have $(2, 2)$ but we do not have $(-2, -2)$ on the graph.

The graph is

Therefore, the graph has no symmetry.

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