# Math 105 Exam 1 Practice (old)

## 1. Compute and simplify.

### (a) $\dfrac{3}{7} \cdot \left(-\dfrac{5}{2}\right)$

:::spoiler
<summary> Solution:</summary>

\begin{align*}
\dfrac{3}{7} \cdot \left(-\dfrac{5}{2}\right)&=\dfrac{(3)(-5)}{(7)(2)} \\
&=\dfrac{-15}{14} \\
&=-\dfrac{15}{14}
\end{align*}
:::

### (b) $\dfrac{1}{4}-\dfrac{2}{3}$

:::spoiler
<summary> Solution:</summary>

\begin{align*}
\dfrac{1}{4}-\dfrac{2}{3}&=\dfrac{3}{12}-\dfrac{8}{12} \\
&=\dfrac{3-8}{12} \\
&=\dfrac{-5}{12} \\
&=-\dfrac{5}{12}
\end{align*}
:::

## 2. Calculate: $-4 \cdot 3^2 +2 \cdot 3 - 6$

:::spoiler
<summary> Solution:</summary>

Following Order of Operations (PEMDAS):

| Step | Operation |
| -------- | -------- |
| $-4 \cdot 3^2 +2 \cdot 3 - 6$ | Original Expression |
| $-4 \cdot (9) +2 \cdot 3 - 6$ | Exponents. $3^2=3 \cdot 3=9$ |
| $-36 +2 \cdot 3 - 6$ | Multiplication $(-4)(9)=-36$ |
| $-36 +6 - 6$ | Multiplication $(2)(3)=6$ |
| $-30 - 6$ | Addition $-36+6=-30$ |
| $-36$ | Subtraction $-30-6=-36$ |

:::

## 3. Graph and label the points $(2,5)$, $(-4,5)$, and $(-2,-6)$.


:::spoiler
<summary> Solution:</summary>

The legend for this graph is below.
| Label | Point |
| -------- | -------- |
| $A$ | $(2,5)$ |
| $B$ | $(-4,5)$ |
| $C$ | $(-2,-6)$ |

:::

## 4. Find the distance between the pair of points $(2,-4)$ and $(5,1)$. Express your answer in the form $\sqrt{N}$, where $N$ is a whole number.

:::spoiler
<summary> Solution:</summary>

\begin{align*}
d&=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\
&=\sqrt{(5-2)^2+(1-(-4))^2} \\
&=\sqrt{3^2+5^2} \\
&=\sqrt{9+25} \\
&=\sqrt{34}
\end{align*}
:::

## 5. Find the midpoint of the segment having the endpoints $\left(\dfrac{2}{3},\dfrac{10}{7}\right)$, $\left(\dfrac{9}{2},-\dfrac{2}{5}\right)$. Your answer should have the form $\left(\dfrac{a}{b},\dfrac{c}{d}\right)$.

:::spoiler
<summary> Solution:</summary>

\begin{align*}
x_m&=\dfrac{x_1+x_2}{2} \\
&=\dfrac{\dfrac{2}{3}+\dfrac{9}{2}}{2} \\
&=\dfrac{\dfrac{4}{6}+\dfrac{27}{6}}{2} \\
&=\dfrac{\dfrac{4+27}{6}}{2} \\
&=\dfrac{\dfrac{31}{6}}{2} \\
&=\dfrac{31}{6} \cdot \dfrac{1}{2} \\
&=\dfrac{31}{12}
\end{align*}

---
\begin{align*}
y_m&=\dfrac{y_1+y_2}{2} \\
&=\dfrac{\dfrac{10}{7}+\dfrac{-2}{5}}{2} \\
&=\dfrac{\dfrac{50}{35}+\dfrac{-14}{35}}{2} \\
&=\dfrac{\dfrac{50+(-14)}{35}}{2} \\
&=\dfrac{\dfrac{36}{35}}{2} \\
&=\dfrac{36}{35} \cdot \dfrac{1}{2} \\
&=\dfrac{18}{35} \cdot \dfrac{1}{1} \\
&=\dfrac{18}{35}
\end{align*}

---
\begin{align*}
\text{Midpoint}&=\left(\dfrac{31}{12},\dfrac{18}{35}\right)
\end{align*}
:::

## 6. Is $\left(\dfrac{3}{2},\dfrac{4}{3}\right)$ a solution to the equation $8x-6y=5$? Show your work.

:::spoiler

<summary> Solution: </summary>

**Understanding the Verification Process**

Let's walk through a detailed check of whether the points $(\frac{3}{2}, \frac{4}{3})$ satisfy the equation $8x - 6y = 5$. We'll break down each part of the process to make it as easy to follow as possible.

1. **Start with the Original Equation**

We're working with the equation $8x - 6y = 5$. This will be our reference point.

2. **Insert the $x$ Value**

Replace $x$ in the equation with $\frac{3}{2}$. This gives us a new equation to work with: $8(\frac{3}{2}) - 6y = 5$.

3. **Insert the $y$ Value**

Now, let's plug in $\frac{4}{3}$ for $y$. After doing so, our equation looks like this: $8(\frac{3}{2}) - 6(\frac{4}{3}) = 5$.

4. **Simplify the Multiplication**

a. First, focus on the $x$ part of the equation. The operation we need to perform is $8 \times \frac{3}{2}$. This can be expressed in fraction form as $\frac{8}{1} \cdot \frac{3}{2}$. To simplify, we can first divide both the numerator and the denominator of the first fraction by $2$, which gives us $\frac{4}{1} \cdot \frac{3}{1}$. Multiplying these simplified fractions together, we obtain $12$.

b. Next, tackle the $y$ part of the equation. We need to perform the operation $6 \times \frac{4}{3}$. This can also be expressed in fraction form as $\frac{6}{1} \times \frac{4}{3}$. To simplify, we can first divide both the numerator of the second fraction and the denominator of the first fraction by $3$, which gives us $\frac{2}{1} \times \frac{4}{1}$. Multiplying these simplified fractions together, we obtain $8$.

5. **Subtract and Simplify**

Replace the multiplication results back into the equation, so it now reads: $12 - 8 = 5$.

6. **Perform the Final Calculation**

Subtract $8$ from $12$, which gives us $4$. So, our simplified equation is now $4 = 5$.

7. **Reach a Conclusion**

Since $4$ does not equal $5$, we find that there's a discrepancy. This tells us that the points $(\frac{3}{2}, \frac{4}{3})$ do not satisfy the original equation $8x - 6y = 5$.

:::

## 7. Find the intercepts and graph the line $$2x-4y=8$$


:::spoiler
<summary> Solution:</summary>

To find the intercepts and graph the line given by the equation $2x - 4y = 8$, we will follow these steps:

1. **Find the x-intercept:** To find the x-intercept, we set $y = 0$ and solve for $x$.
2. **Find the y-intercept:** To find the y-intercept, we set $x=0$ and solve for $y$.
3. **Graph the line:** Using the intercepts found in steps 1 and 2, we will plot the line on a coordinate plane.

Let's start by finding the intercepts.

### Step 1: Find the x-intercept

To find the x-intercept, set $y=0$ in the equation $2x - 4y = 8$, and solve for $x$.

$$2x - 4(0) = 8$$
$$2x = 8$$
$$x = 4$$

Thus, the x-intercept is at $(4, 0)$.

### Step 2: Find the y-intercept

To find the y-intercept, set $x=0$ in the equation $2x - 4y = 8$, and solve for $y$.

$$2(0) - 4y = 8$$
$$-4y = 8$$
$$y = -2$$

Thus, the y-intercept is at $(0, -2)$.

### Step 3: Graph the line

Now, with the intercepts $(4, 0)$ and $(0, -2)$, we can graph the line. We'll plot these points on a coordinate plane and draw a line through them.

Here is the graph of the line $2x - 4y = 8$ using the intercepts we found:


- The **x-intercept** is at $(4, 0)$, marked with a red dot on the graph.
- The **y-intercept** is at $(0, -2)$, also marked with a red dot on the graph.

As you can see, the line passes through these intercepts, illustrating how the equation $2x - 4y = 8$ is represented visually on a coordinate plane.
:::

## 8. Is $\{(0,4),(-1,6),(1,6),(2,6), (-2,3), (2,5)\}$ a function? Explain.

:::spoiler
<summary> Solution:</summary>

### Is the Set a Function?

To determine if the set $\{(0,4),(-1,6),(1,6),(2,6), (-2,3), (2,5)\}$ is a function, we need to examine if every input (or x-value) maps to exactly one output (or y-value).

A function, by definition, assigns exactly one output to each input. In other words, an x-value cannot correspond to more than one y-value.

Let's examine the given set:

$$
\{(0,4),(-1,6),(1,6),(2,6), (-2,3), (2,5)\}
$$

In this set, each input (x-value) appears to be paired with only one output (y-value) with one exception: the input $2$ is paired with both $6$ and $5$ (as seen in the pairs $(2,6)$ and $(2,5)$).

Because the input $2$ corresponds to more than one output, the given set **does not represent a function**. For a set of ordered pairs to be a function, each unique input must map to exactly one output, which is not the case here.

The set/relation can also be represented by the following table:

| x | y |
|---|---|
| 0 | 4 |
| -1| 6 |
| 1 | 6 |
| 2 | 6 |
| -2| 3 |
| 2 | 5 |

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## 9. A graph of a function $f(x)$ is shown in Figure 1. Using the graph, find the value of $f(1)$, $f(-2)$, and $f(0)$.


:::spoiler
<summary> Solution:</summary>

The graph passes through the point $(1,-1)$, thus $f(1)=-1$.

The graph passes through the point $(-2,1)$, thus $f(-2)=1$.

The graph passes through the point $(0,3)$, thus $f(0)=3$.

:::

## 10. Using the same graph from Figure 1, find the domain and range of the function using interval notation.

:::spoiler
<summary> Solution:</summary>

### Finding the Domain and Range from a Graph

Given the graph of a function $y = f(x)$, we aim to determine the domain and range of the function.

#### Domain
The domain of a function consists of all the input values (x-values) for which the function is defined. Based on the graph provided and assuming the graph depicts the entire function:

$$
\text{Domain: }[-3, 3]
$$

This indicates that the function $f(x)$ is defined for all values of $x$ from -3 to 3, inclusive. In inequality notation this is $-3 \leq x \leq 3$.

#### Range
The range of a function includes all the output values (y-values) that the function can produce. Again, based on the provided graph and the given assumption:

$$
\text{Range: } [-1, 3]
$$

This tells us that the function $f(x)$ produces values of $y$ from -1 to 3, inclusive. In inequality notation this is $-1 \leq y \leq 3$.

The square brackets $[ ]$ denote that the endpoints are included in the domain and range, which means the function includes the values at $x = -3$ and $x = 3$ for the domain, and $y = -1$ and $y = 3$ for the range.

:::

## 11. Determine the domain of the function $$f(x)=\dfrac{x+5}{7-x}$$

:::spoiler
<summary> Solution:</summary>
The domain of a function is the set of all possible input values (x-values) for which the function is defined. In the case of the function

$$
f(x)=\dfrac{x+5}{7-x}
$$

we need to consider any restrictions on the domain. The only restriction for this rational function comes from the denominator, since division by zero is undefined. Therefore, we need to find the value of $x$ that would make the denominator zero and exclude it from the domain.

Setting the denominator equal to zero, we solve for $x$:

$$
7 - x = 0
$$
$$
x = 7
$$

Since $x=7$ would result in division by zero, it must be excluded from the domain. Thus, the domain of $f(x)$ is all real numbers except $x = 7$.

Using interval notation, the domain of $f(x)$ is:

$$
(-\infty, 7) \cup (7, +\infty)
$$

This represents all real numbers less than 7 and all real numbers greater than 7.
:::

## 12. Let $$f(x)=\dfrac{x}{x-3}.$$ Find $$f(3)=$$ $$f(6)=$$ $$f\left(\dfrac{1}{2}\right)=$$

:::spoiler
<summary> Solution:</summary>
The function given is:

$$
f(x) = \frac{x}{x-3}
$$

Now let's evaluate the function for the specified values of x.

For $f(3)$:

When $x = 3$, the denominator of the function $x-3$ becomes zero, which means the function is undefined. So, $f(3)$ does not exist because it would require division by zero. Verify this for yourself.

> I can only show you the door. You're the one that has to walk through it
> -Morpheus, __The Matrix__

For $f(6)$:

When $x = 6$:

$$
f(6) = \frac{6}{6-3} = \frac{6}{3} = 2
$$

For $f\left(\frac{1}{2}\right)$:

When $x = \frac{1}{2}$:

$$
f\left(\frac{1}{2}\right) = \frac{\frac{1}{2}}{\frac{1}{2}-3} = \frac{\frac{1}{2}}{-\frac{5}{2}} = \frac{1}{2} \cdot \frac{-2}{5} = \dfrac{-2}{10}=-\frac{1}{5}
$$

(Note that for the denominator, we combine the fractions: $\dfrac{1}{2}-3=\dfrac{1}{2}-\dfrac{3}{1}=\dfrac{1}{2}-\dfrac{6}{2}=\dfrac{-5}{2}$)

Therefore, the calculated values are:

- $f(3)$ is undefined.
- $f(6) = 2$
- $f\left(\frac{1}{2}\right) = -\frac{1}{5}$

:::

## 13. Consider the graph in Figure 2. Is this a function? Explain briefly.


:::spoiler
<summary> Solution:</summary>


### Determining If a Graph Represents a Function Using the Vertical Line Test

When analyzing a graph to determine if it represents a function, one crucial tool we use is the Vertical Line Test. A graph represents a function if and only if no vertical line can intersect the graph at more than one point.

## Problem Description

Given a graph of a curve that loops back over itself, we find that it's possible to draw a vertical line that intersects the graph at three distinct points.

## Applying the Vertical Line Test

To apply the Vertical Line Test:

1. **Visualize or Draw a Vertical Line**: Imagine drawing a vertical line anywhere on the graph.
2. **Count the Intersection Points**: Count how many times the vertical line intersects the graph.
- In this case, we have identified a spot where a vertical line intersects the graph at three points.
3. **Analyze the Results**: According to the Vertical Line Test, if a vertical line intersects the graph more than once, then the graph does not represent a function.

## Conclusion

Since there is at least one vertical line that intersects our given graph at three points, we conclude that **the graph does not represent a function**. This is because a function, by definition, assigns exactly one output (y-value) for each input (x-value), which is violated in this scenario.

This understanding is crucial for distinguishing between graphs that represent functions and those that do not, ensuring clarity in the definition and representation of mathematical functions.

---
Extra
---
The Vertical Line Test is a visual method used to determine whether a graph represents a function. This test is based on the definition of a function in mathematics, where each input (or x-value) is associated with exactly one output (or y-value). Let's delve deeper into how the Vertical Line Test elucidates this definition and its implications.

## The Principle Behind the Vertical Line Test

The core principle of the Vertical Line Test is rooted in the definition of a function:

- **A Function**: For a set of points in a Cartesian coordinate system to represent a function, every x-value must correspond to exactly one y-value. This is a foundational characteristic of functions—it ensures that for each input, there is a unique output.

- **Non-Function Graphs**: If a graph includes any x-value that corresponds to more than one y-value, it does not represent a function. This can occur in various types of graphs, such as circles, ellipses, or any curve that loops back over itself.

## Applying the Vertical Line Test

- **Intersecting Once**: When you draw a vertical line (parallel to the y-axis) anywhere on the graph, and it intersects the graph at exactly one point, it indicates that for that particular x-value, there is only one y-value. If this is true for every possible vertical line you could draw (covering the entire domain of the graph), the graph passes the Vertical Line Test, confirming it represents a function.

- **Intersecting More Than Once**: Conversely, if you can draw a vertical line that intersects the graph at more than one point, it demonstrates that there exists an x-value with multiple y-values. This scenario violates the fundamental definition of a function. Even if just one vertical line intersects the graph multiple times, the graph fails the Vertical Line Test, indicating it does not represent a function.

## Why the Test Matters

The Vertical Line Test offers a quick and intuitive way to assess the fundamental property of functions regarding input-output uniqueness. It helps visually distinguish between graphs that adhere to the definition of a function and those that do not, without having to delve into the underlying equations.

## Examples and Exceptions

- **Functions**: Graphs of linear equations, parabolas, or any curve where each x-value maps to a single y-value.
- **Non-Functions**: Circles, vertical lines (except the y-axis itself), or any figure where a loop or curve allows for a vertical line to cut through the graph at multiple points.

In essence, the Vertical Line Test embodies the concept of functional uniqueness and provides a straightforward graphical tool to verify the fundamental criterion that defines functions in mathematics.

:::

## 14. Find the slope of the line through the points $(2,6)$ and $(-8,-2)$.

:::spoiler
<summary> Solution:</summary>

To find the slope of the line that passes through the points $(2,6)$ and $(-8,-2)$, we use the slope formula:

$$
m = \frac{y_2 - y_1}{x_2 - x_1}
$$

where $(x_1, y_1)$ and $(x_2, y_2)$ are the coordinates of the two points.

Substituting the given points into the formula gives us:

$$
m = \frac{-2 - 6}{-8 - 2} = \frac{-8}{-10} = \frac{4}{5}
$$

Therefore, the slope of the line through the points $(2,6)$ and $(-8,-2)$ is $\frac{4}{5}$.


:::

## 15. Determine the slope and intercepts of the line or state that they do not exist: $7x-3y=5$.

:::spoiler
<summary> Solution:</summary>

To determine the slope and intercepts of the line given by the equation $7x - 3y = 5$, we can rewrite the equation in slope-intercept form, which is $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

Starting with the given equation:

$$
7x - 3y = 5
$$

We solve for $y$ to put it into slope-intercept form:

$$
-3y = -7x + 5
$$

$$
y = \frac{7}{3}x - \frac{5}{3}
$$

Now, we can identify the slope and y-intercept:

- The slope ($m$) is $\frac{7}{3}$.
- The y-intercept ($b$) is $-\frac{5}{3}$, which means the line crosses the y-axis at $(0, -\frac{5}{3})$.

To find the x-intercept, we set $y = 0$ and solve for $x$:

$$
7x - 3(0) = 5
$$

$$
7x = 5
$$

$$
x = \frac{5}{7}
$$

So the x-intercept is $\left(\frac{5}{7}, 0\right)$.

Therefore, the slope of the line is $\frac{7}{3}$, the y-intercept is $(0, -\frac{5}{3})$, and the x-intercept is $\left(\frac{5}{7}, 0\right)$.

:::

## Formulas

$$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

$$\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$$

$$\dfrac{y_2-y_1}{x_2-x_1}$$