# Math 105 Exam 1b Practice

## 1. Compute and simplify.

### (a) $\dfrac{5}{6} \cdot \left(-\dfrac{7}{3}\right)$

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<summary> Solution:</summary>

\begin{align*}
\dfrac{5}{6} \cdot \left(-\dfrac{7}{3}\right)&=\dfrac{(5)(-7)}{(6)(3)} \\
&=\dfrac{-35}{18} \\
&=-\dfrac{35}{18}
\end{align*}
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### (b) $\dfrac{2}{5}-\dfrac{1}{4}$

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<summary> Solution:</summary>

\begin{align*}
\dfrac{2}{5}-\dfrac{1}{4}&=\dfrac{8}{20}-\dfrac{5}{20} \\
&=\dfrac{8-5}{20} \\
&=\dfrac{3}{20}
\end{align*}
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## 2. Simplify: $-5 \cdot (-3)^2 - 4 \cdot (-2) - 7$

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

Following Order of Operations (PEMDAS):

| Step | Operation |
| -------- | -------- |
| $-5 \cdot (-3)^2 - 4 \cdot (-2) - 7$ | Original Expression |
| $-5 \cdot (9) - 4 \cdot (-2) - 7$ | Exponents. $(-3)^2=(-3) \cdot (-3)=9$ |
| $-45 - 4 \cdot (-2) - 7$ | Multiplication $(-5)(9)=-45$ |
| $-45 + 8 - 7$ | Multiplication $(-4)(-2)=8$ |
| $-37 - 7$ | Addition $-45 + 8 = -37$ |
| $-44$ | Subtraction $-37 - 7 = -44$ |

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## 3. Graph and label the points $(3,-2)$, $(-5,-4)$, and $(-6,1)$.


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<summary> Solution:</summary>


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## 4. Find the distance between the pair of points $(-3,7)$ and $(4,-2)$. 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{(4-(-3))^2+(-2-7)^2} \\
&=\sqrt{(4+3)^2+(-9)^2} \\
&=\sqrt{7^2+(-9)^2} \\
&=\sqrt{49+81} \\
&=\sqrt{130}
\end{align*}
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## 5. Find the midpoint of the segment having the endpoints $\left(6,\dfrac{3}{4}\right)$, $\left(-2,-\dfrac{5}{6}\right)$. Your answer should have the form $\left(\dfrac{a}{b},\dfrac{c}{d}\right)$.

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<summary> Solution:</summary>

\begin{align*}
x_m&=\dfrac{x_1+x_2}{2} \\
&=\dfrac{6+(-2)}{2} \\
&=\dfrac{4}{2} \\
&=2
\end{align*}

---
\begin{align*}
y_m&=\dfrac{y_1+y_2}{2} \\
&=\dfrac{\dfrac{3}{4}+\dfrac{-5}{6}}{2} \\
&=\dfrac{\dfrac{9}{12}+\dfrac{-10}{12}}{2} \\
&=\dfrac{\dfrac{-1}{12}}{2} \\
&=\dfrac{-1}{12} \cdot \dfrac{1}{2} \\
&=\dfrac{-1}{24}
\end{align*}

---
\begin{align*}
\text{Midpoint}&=\left(2,-\dfrac{1}{24}\right)
\end{align*}
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## 6. Is $\left(\dfrac{5}{3}, \dfrac{7}{2}\right)$ a solution to the equation $9x + 4y = 23$? Show your work.

:::spoiler

<summary> Solution: </summary>

To determine if the point $\left(\frac{5}{3}, \frac{7}{2}\right)$ is a solution to the equation $9x + 4y = 23$, we need to substitute $x = \frac{5}{3}$ and $y = \frac{7}{2}$ into the equation and check if it holds true.

The original equation is:

$9x + 4y = 23$

### Step 1: Substitute $x = \frac{5}{3}$ and $y = \frac{7}{2}$ into the equation.

$9\left(\frac{5}{3}\right) + 4\left(\frac{7}{2}\right) = 23$

### Step 2: Simplify each term.

$9 \times \frac{5}{3} = \frac{9 \times 5}{3} = 15$

$4 \times \frac{7}{2} = \frac{4 \times 7}{2} = 14$

### Step 3: Substitute the simplified terms back into the equation.

$15 + 14 = 23$

### Step 4: Simplify the result.

$29 \neq 23$

Since $29 \neq 23$, the point $\left(\frac{5}{3}, \frac{7}{2}\right)$ is **not** a solution to the equation $9x + 4y = 23$.

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## 7. Find the intercepts and graph the line $$3x-5y=15$$


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

To find the intercepts and graph the line given by the equation $3x - 5y = 15$, 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 $3x - 5y = 15$, and solve for $x$.

$$3x - 5(0) = 15$$
$$3x = 15$$
$$x = 5$$

Thus, the x-intercept is at $(5, 0)$.

### Step 2: Find the y-intercept

To find the y-intercept, set $x = 0$ in the equation $3x - 5y = 15$, and solve for $y$.

$$3(0) - 5y = 15$$
$$-5y = 15$$
$$y = -3$$

Thus, the y-intercept is at $(0, -3)$.

### Step 3: Graph the line

Now, with the intercepts $(5, 0)$ and $(0, -3)$, 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 $3x - 5y = 15$ using the intercepts we found:


- The **x-intercept** is at $(5, 0)$, marked with a red dot on the graph.
- The **y-intercept** is at $(0, -3)$, also marked with a red dot on the graph.

As you can see, the line passes through these intercepts, illustrating how the equation $3x - 5y = 15$ is represented visually on a coordinate plane.

:::

## 8. Is this a function? Explain.

| domain | range |
|--------|-------|
| 3 | 5 |
| -2 | 5 |
| -1 | 7 |
| 0 | 4 |
| 1 | 5 |
| 2 | 7 |
| 3 | 8 |
| 4 | 9 |

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

To determine if the given relation is a function, we need to check if each element in the domain corresponds to exactly one element in the range.

### Explanation:
For the relation to be a function, each value in the domain must map to only one value in the range.

- In this table, the domain value $3$ is mapped to both $5$ and $8$.
- This violates the definition of a function, which states that each input (domain) must have exactly one output (range).

### Conclusion:
Since the domain value $3$ has two different range values, this relation is **not a function**.

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## 9. Is this a function? Explain.


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

To determine if the relation represented by the diagram defines a function, we must check if every element in the domain (left side) is mapped to **exactly one** element in the range (right side).

#### Definition of a Function:
A relation is a function if **each input (domain)** has **exactly one output (range)**. If any input corresponds to more than one output, the relation is **not** a function.

#### Explanation of the Diagram:
- The domain consists of the elements \( \{1, 2, 3, 4\} \).
- The range consists of the elements \( \{A, B, C, D\} \).
- The mappings are as follows:
- 1 is mapped to B.
- 2 is mapped to C.
- 3 is mapped to D.
- 4 is mapped to A.

#### Conclusion:
In the diagram:
- Each element in the domain \( \{1, 2, 3, 4\} \) is mapped to exactly **one** element in the range.
- No domain element has more than one arrow leading out of it, and no domain element is left unmapped.

Since every element in the domain has exactly one corresponding element in the range, **this relation is a function**.

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## 10. Is {(3, 8), (4, 7), (-1, 5), (3, 9), (0, 6)} a function? Explain.

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

To determine if the relation $\{(3, 8), (4, 7), (-1, 5), (3, 9), (0, 6)\}$ is a function, we need to check whether each input (domain value) corresponds to exactly one output (range value).

### Definition:
A relation is a function if every element in the domain has exactly one corresponding element in the range. This means no domain value can map to more than one range value.

### Domain and Range:
- The given relation contains the following pairs: $\{(3, 8), (4, 7), (-1, 5), (3, 9), (0, 6)\}$.
- The domain values are: $3, 4, -1, 3, 0$.
- The range values are: $8, 7, 5, 9, 6$.

### Analysis:
- The domain value $3$ appears twice in the set: $(3, 8)$ and $(3, 9)$.
- This means the domain value $3$ is mapped to two different range values ($8$ and $9$), which violates the definition of a function.

### Conclusion:
Since the domain value $3$ is associated with two different range values ($8$ and $9$), this relation is **not a function**.

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


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

$f(1)=3$ since the graph passes through the point $(1,3)$

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

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

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## 12. Using the same graph from Figure 1 above, 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 the 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:

$$\text{Domain: } [-1, 4]$$

This indicates that the function $f(x)$ is defined for all values of $x$ from -1 to 4, inclusive. In inequality notation, this is $-1 \leq x \leq 4$.

#### Range
The range of a function includes all the output values (y-values) that the function can produce. Based on the graph:

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

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, meaning the function includes the values at $x = -1$ and $x = 4$ for the domain, and $y = -1$ and $y = 3$ for the range.

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## 13. Determine the domain of the function $$f(x)=\dfrac{x-2}{4x+8}$$

:::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

$$
f(x)=\dfrac{x-2}{4x+8}
$$

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 $x$:

$$
4x + 8 = 0
$$
$$
4x = -8
$$
$$
x = -2
$$

Since $x=-2$ would result in division by zero, it must be excluded from the domain. Thus, the domain of $f(x)$ is all real numbers except $x = -2$.

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

$$
(-\infty, -2) \cup (-2, +\infty)
$$

This represents all real numbers less than $-2$ and all real numbers greater than $-2$.

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## 14. Let $$f(x)=\dfrac{4x-5}{x-4}.$$ Find $$f(4)=$$ $$f(0)=$$ $$f\left(\dfrac{2}{3}\right)=$$

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

$$
f(x) = \frac{4x - 5}{x - 4}
$$

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

### For $f(4)$:

When $x = 4$, the denominator of the function $x - 4$ becomes zero, which means the function is undefined. So, $f(4)$ does not exist because it would require division by zero.

### For $f(0)$:

When $x = 0$:

$$
f(0) = \frac{4(0) - 5}{0 - 4} = \frac{-5}{-4} = \frac{5}{4}
$$

### For $f\left(\frac{2}{3}\right)$:

When $x = \frac{2}{3}$:

\begin{align*}
f\left(\frac{2}{3}\right) &= \frac{4\left(\frac{2}{3}\right) - 5}{\frac{2}{3} - 4}
\end{align*}

### For the numerator:
$$
\begin{align*}
\text{Numerator} &= 4\left(\frac{2}{3}\right) - 5 \\
&= \frac{8}{3} - 5 \\
&= \frac{8}{3} - \frac{15}{3} \\
&= \frac{-7}{3}
\end{align*}
$$

### For the denominator:
$$
\begin{align*}
\text{Denominator} &= \frac{2}{3} - 4 \\
&= \frac{2}{3} - \frac{12}{3} \\
&= \frac{-10}{3}
\end{align*}
$$

### Combine numerator and denominator:
$$
\begin{align*}
f\left(\frac{2}{3}\right) &= \frac{\frac{-7}{3}}{\frac{-10}{3}} \\
&= \frac{-7}{3} \times \frac{3}{-10} \\
&= \frac{7}{10}
\end{align*}
$$

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## 15a. 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.

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## 15b. Consider the graph in Figure 2. Is this a function? Explain briefly.


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


To determine if the graph represents a function, we apply the **Vertical Line Test**.

### Vertical Line Test Explanation:
- A graph represents a function if **no vertical line** intersects the graph at more than one point.
- If any vertical line crosses the graph more than once, the graph does **not** represent a function.

### Analyzing the Given Graph:
The provided graph appears to be a **parabola** opening upwards, specifically resembling the curve of $y = x^2$. This graph:
- **Passes the Vertical Line Test**: No matter where a vertical line is drawn, it will intersect the graph at **only one point** for each value of $x$.
- Since each input $x$-value corresponds to **exactly one output $y$-value**, the graph satisfies the definition of a function.

### Conclusion:
The graph **does** represent a function because it passes the Vertical Line Test. For each $x$-value, there is only one corresponding $y$-value. Therefore, the graph is a valid function.

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## 16. Determine the slope of the line that passes through the points $(-3, -9)$ and $(-5, -7)$.

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

To find the slope of the line passing through the points $(-3, -9)$ and $(-5, -7)$, 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:

$$
m = \frac{-7 - (-9)}{-5 - (-3)} = \frac{2}{-2} = -1
$$

Thus, the slope of the line through the points $(-3, -9)$ and $(-5, -7)$ is $-1$.

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## 17. Determine the slope and intercepts of the line or state that they do not exist: $5x - 4y = 9$.

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

To find the slope and intercepts of the line given by the equation $5x - 4y = 9$, we first 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:

$$
5x - 4y = 9
$$

We solve for $y$ by first subtracting $5x$ from both sides and then dividing by $-4$:

\begin{align}
5x - 4y &= 9 \\
-4y &= -5x + 9 \\
\dfrac{-4y}{-4} &= \dfrac{-5x + 9}{-4} \\
y &= \frac{5}{4}x - \frac{9}{4}
\end{align}

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

- The slope ($m$) is $\frac{5}{4}$.
- The y-intercept ($b$) is $-\frac{9}{4}$, so the line crosses the y-axis at $(0, -\frac{9}{4})$.

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

$$
5x - 4(0) = 9
$$

$$
5x = 9
$$

$$
x = \frac{9}{5}
$$

Thus, the x-intercept is $\left(\frac{9}{5}, 0\right)$.

Therefore, the slope of the line is $\frac{5}{4}$, the y-intercept is $(0, -\frac{9}{4})$, and the x-intercept is $\left(\frac{9}{5}, 0\right)$.

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## 18. Determine the slope and intercepts of the line or state that they do not exist: $y=-2$.

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

$y=-2$ is a horizontal line that passes through its $y$-intercept $(0,-2)$. Horizontal lines have zero slope ($m=0$). It has no $x$-intercept.


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## Formulas

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

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

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