# Math 105 Quiz 1

Work in groups and submit one submission with everyone's names on it. Make sure everyone scans the paper to PDF and shares it with the group.

## 1. Compute and simplify.

### (a) $\dfrac{7}{5} \cdot \left(-\dfrac{4}{9}\right)$

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<summary> Solution: </summary>

\begin{align*}
\dfrac{7}{5} \cdot \left(-\dfrac{4}{9}\right) &= \dfrac{7 \cdot (-4)}{5 \cdot 9} \\
&= \dfrac{-28}{45} \\
&=-\dfrac{28}{45}
\end{align*}

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### (b) $\dfrac{1}{3}-\dfrac{5}{8}$

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<summary> Solution: </summary>

\begin{align*}
\dfrac{1}{3} - \dfrac{5}{8} &= \dfrac{8}{24} - \dfrac{15}{24} \\
&= \dfrac{8 - 15}{24} \\
&= \dfrac{-7}{24}
\end{align*}

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## 2. Simplify: $-4 \cdot (-5)^2 - 3 \cdot (-4) - 6$

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<summary> Solution: </summary>

\begin{align*}
-4 \cdot (-5)^2 - 3 \cdot (-4) - 6 &= -4 \cdot 25 - 3 \cdot (-4) - 6 \\
&= -100 + 12 - 6 \\
&= -100 + 6 \\
&= -94
\end{align*}

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

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<summary> Solution: </summary>


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## 4. Find the distance between the pair of points $(-2,6)$ and $(3,-1)$. Express your answer in the form $\sqrt{N}$, where $N$ is a whole number.

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<summary> Solution: </summary>

\begin{align*}
d &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\
&= \sqrt{(3 - (-2))^2 + (-1 - 6)^2} \\
&= \sqrt{(3 + 2)^2 + (-1 - 6)^2} \\
&= \sqrt{5^2 + (-7)^2} \\
&= \sqrt{25 + 49} \\
&= \sqrt{74}
\end{align*}

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## 5. Find the midpoint of the segment having the endpoints $\left(3,\dfrac{2}{5}\right)$, $\left(-5,-\dfrac{1}{6}\right)$. Your answer should have the form $\left(\dfrac{a}{b},\dfrac{c}{d}\right)$.

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<summary> Solution: </summary>

\begin{align*}
x_{\text{mid}} &= \dfrac{x_1 + x_2}{2} \\
&= \dfrac{3 + (-5)}{2} \\
&= \dfrac{3 - 5}{2} \\
&= \dfrac{-2}{2} \\
&= -1
\end{align*}

\begin{align*}
y_{\text{mid}} &= \dfrac{y_1 + y_2}{2} \\
&= \dfrac{\dfrac{2}{5} + \left(-\dfrac{1}{6}\right)}{2} \\
&= \dfrac{\dfrac{2}{5} - \dfrac{1}{6}}{2} \\
&= \dfrac{\dfrac{12}{30} - \dfrac{5}{30}}{2} \\
&= \dfrac{\dfrac{7}{30}}{2} \\
&= \dfrac{7}{30} \div \dfrac{2}{1} \\
&= \dfrac{7}{30} \cdot \dfrac{1}{2} \\
&= \dfrac{7}{60}
\end{align*}

$$\text{Midpoint}=\left( -1, \dfrac{7}{60} \right)$$

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## 6. Is $\left(\dfrac{2}{5}, \dfrac{3}{4}\right)$ a solution to the equation $10x + 8y = 10$? Show your work.

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<summary> Solution: </summary>

To determine if the point $\left(\dfrac{2}{5}, \dfrac{3}{4}\right)$ is a solution to the equation $10x + 8y = 10$, we substitute $x = \dfrac{2}{5}$ and $y = \dfrac{3}{4}$ into the equation:

$$
10\left(\dfrac{2}{5}\right) + 8\left(\dfrac{3}{4}\right)
$$

### Step-by-Step Calculation:

1. **First term:**

$$
10\left(\dfrac{2}{5}\right) =\dfrac{10}{1} \cdot \dfrac{2}{5}= \dfrac{10 \cdot 2}{5} = \dfrac{20}{5} = 4
$$

2. **Second term:**

$$8\left(\dfrac{3}{4}\right) =\dfrac{8}{1} \cdot \dfrac{3}{4} =\dfrac{8 \cdot 3}{4} = \dfrac{24}{4} = 6$$

3. **Adding the two terms:**

$$
4 + 6 = 10
$$

### Conclusion:
Since the left-hand side equals the right-hand side (both are 10), $\left(\dfrac{2}{5}, \dfrac{3}{4}\right)$ is **a solution** to the equation $10x + 8y = 10$.

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

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<summary> Solution: </summary>

### 1. Finding the x-intercept:
The x-intercept occurs when $y = 0$. To find the x-intercept, set $y = 0$ and solve for $x$:

$$
\begin{align*}
4x + 2(0) &= 8 \\
4x &= 8 \\
x &= \dfrac{8}{4} \\
x &= 2
\end{align*}
$$

So, the x-intercept is $(2, 0)$.

### 2. Finding the y-intercept:
The y-intercept occurs when $x = 0$. To find the y-intercept, set $x = 0$ and solve for $y$:

$$
\begin{align*}
4(0) + 2y &= 8 \\
2y &= 8 \\
y &= \dfrac{8}{2} \\
y &= 4
\end{align*}
$$

So, the y-intercept is $(0, 4)$.

### 3. Graphing the line:
Now we have two points: $(2, 0)$ (the x-intercept) and $(0, 4)$ (the y-intercept). Plot these points on a graph and draw a line through them.

The equation of the line is $4x + 2y = 8$.


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## 8. Is this a function? Explain.

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

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<summary> Solution: </summary>

### Explanation:

Looking at the table, we see that the input $2$ maps to two different outputs: $4$ and $5$. This means that the domain value $2$ corresponds to more than one range value, which violates the definition of a function.

### Conclusion:

Since the input $2$ is associated with two different outputs ($4$ and $5$), **this is not a function**.

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## 9. Is this a function? Explain.


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<summary> Solution: </summary>

### Is this a function? Explain.

To determine if this is a function, we use the definition of a function: **a function is a relation in which each input (domain element) maps to exactly one output (range element)**.

### Conclusion:
This **is a function** because every input maps to one output.

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## 10. Is {(2, 8), (4, 7), (-1, 5), (3, 9), (0, 6)} a function? Explain.

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<summary> Solution: </summary>

To determine if this set of ordered pairs represents a function, we need to check if each input (first element in the pair) maps to exactly one output (second element in the pair).

The given set of ordered pairs is:

$$(2, 8), (4, 7), (-1, 5), (3, 9), (0, 6)$$

### Explanation:
Each input (2, 4, -1, 3, 0) corresponds to exactly one unique output (8, 7, 5, 9, 6). No input is repeated with a different output.

### Conclusion:
Yes, this set **is a function** because each input has only one output.

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## 11. A graph of a function $f(x)$ is shown in Figure 1. Using the graph, find the value of $f(0)$, $f(3)$ and $f(4)$.


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<summary> Solution: </summary>

1. **Find $f(0)$**:
- At $x = 0$, the value of the function corresponds to the point on the graph where $x = 0$.
- From the graph, when $x = 0$, the value of $y$ is $2.5$.
- Therefore, $f(0) = 2.5$.

2. **Find $f(3)$**:
- At $x = 3$, we check the point on the graph where $x = 3$.
- From the graph, when $x = 3$, the value of $y$ is $2$.
- Therefore, $f(3) = 2$.

3. **Find $f(4)$**:
- At $x = 4$, we check the point on the graph where $x = 4$.
- From the graph, when $x = 4$, the value of $y$ is $1$.
- Therefore, $f(4) = 1$.

### Conclusion:
- $f(0) = 2.5$
- $f(3) = 2$
- $f(4) = 1$

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

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<summary> Solution: </summary>

### Domain:
The domain refers to all the possible values of $x$ for which the function is defined.

From the graph:
- The function exists from $x = -1$ to $x = 4$.

Thus, the domain is:

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

### Range:
The range refers to all the possible values of $y$ that the function takes.

From the graph:
- The lowest value of $y$ is $-1$ (at $x = 2$).
- The highest value of $y$ is $3$ (at $x = 1$).

Thus, the range is:

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

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## 13. Determine the domain of the function $$f(x)=\dfrac{x-1}{5x-15}$$

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<summary> Solution: </summary>

To determine the domain of the function:

$$
f(x) = \dfrac{x-1}{5x-15}
$$

we need to find the values of $x$ that make the denominator equal to zero, as the function is undefined at those points.

### Step 1: Set the denominator equal to zero

$$
5x - 15 = 0
$$

### Step 2: Solve for $x$

$$
5x = 15
$$

$$
x = 3
$$

### Domain:
The function is undefined at $x = 3$. Therefore, the domain is all real numbers except $x = 3$:

$$
\text{Domain: } (-\infty, 3) \cup (3, \infty)
$$

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## 14. Let $$f(x)=\dfrac{x-3}{x-5}.$$ Find $$f(5)=$$ $$f(7)=$$ $$f\left(\dfrac{1}{4}\right)=$$

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<summary> Solution: </summary>

Given the function:

$$
f(x) = \dfrac{x-3}{x-5}
$$

### Step 1: Find $f(5)$

Substitute $x = 5$ into the function:

$$
f(5) = \dfrac{5-3}{5-5} = \dfrac{2}{0}
$$

Since division by zero is undefined:

$$
f(5) \text{ is undefined.}
$$

### Step 2: Find $f(7)$

Substitute $x = 7$ into the function:

$$
f(7) = \dfrac{7-3}{7-5} = \dfrac{4}{2} = 2
$$

### Step 3: Find $f\left(\dfrac{1}{4}\right)$

Substitute $x = \dfrac{1}{4}$ into the function:

$$
f\left(\dfrac{1}{4}\right) = \dfrac{\dfrac{1}{4} - 3}{\dfrac{1}{4} - 5}
$$

### Numerator:

We calculate the numerator in detail:

$$
\begin{align}
\text{Numerator: } &=\dfrac{1}{4} - 3 \\
&=\dfrac{1}{4} - \dfrac{3}{1} \\
&= \dfrac{1}{4} - \dfrac{12}{4} \\
&= \dfrac{1 - 12}{4} \\
&= \dfrac{-11}{4}
\end{align}
$$

### Denominator:

Now, we calculate the denominator in detail:

$$
\begin{align}
\text{Denominator: } &=\dfrac{1}{4} - 5 \\
&=\dfrac{1}{4} - \dfrac{5}{1} \\
&= \dfrac{1}{4} - \dfrac{20}{4} \\
&= \dfrac{1 - 20}{4} \\
&= \dfrac{-19}{4}
\end{align}
$$

### Division of Fractions:

Now, divide the fraction by another fraction:

$$
\begin{align}
f\left(\dfrac{1}{4}\right) &= \dfrac{-\dfrac{11}{4}}{-\dfrac{19}{4}} \\
&= \dfrac{-11}{4} \times \dfrac{4}{-19} \quad \text{(multiply by the reciprocal of the denominator)} \\
&= \dfrac{-11 \times 4}{4 \times -19} \\
&= \dfrac{-44}{-76} \\
&= \dfrac{11}{19} \qquad \text{ Divided top and bottom by $-4$}
\end{align}
$$

### Final Result:

$$
f\left(\dfrac{1}{4}\right) = \dfrac{11}{19}
$$

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


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<summary> Solution: </summary>

No, the graph doesn't represent a function. It intersects a vertical line more than once. Thus it fails the vertical line test.


One input $x$ has multiple outputs $y$.

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## 16. Determine the slope of the line that passes through the points $(-2, -5)$ and $(-4, -8)$.

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<summary> Solution: </summary>

To determine the slope of the line that passes through the points $(-2, -5)$ and $(-4, -8)$, we use the slope formula:

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

where:
- $(x_1, y_1) = (-2, -5)$
- $(x_2, y_2) = (-4, -8)$

Now, substitute the values:

$$
m = \dfrac{-8 - (-5)}{-4 - (-2)}
$$

Simplify:

$$
m = \dfrac{-8 + 5}{-4 + 2} = \dfrac{-3}{-2} = \dfrac{3}{2}
$$

Thus, the slope of the line is:

$$
m = \dfrac{3}{2}
$$

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## 17. Determine the slope and intercepts of the line or state that they do not exist: $2x - 5y = 3$.

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<summary> Solution: </summary>

Given the equation of the line:

$$
2x - 5y = 3
$$

### Step 1: Find the slope and $y$-intercept

To find the slope and $y$-intercept, we need to rewrite the equation in slope-intercept form ($y = mx + b$). Start by solving for $y$:

$$
\begin{align}
2x - 5y &= 3 \\
-5y &= -2x + 3 \\
y &= \dfrac{-2x + 3}{-5} \\
y &= \dfrac{2}{5}x - \dfrac{3}{5}
\end{align}
$$

Now, the equation is in slope-intercept form:

$$
y = \dfrac{2}{5}x - \dfrac{3}{5}
$$

From this, we can see that:

- **Slope** ($m$): $\dfrac{2}{5}$
- **$y$-intercept** ($b$): $-\dfrac{3}{5}$

### Step 2: Find the $x$-intercept

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

$$
\begin{align}
2x - 5(0) &= 3 \\
2x &= 3 \\
x &= \dfrac{3}{2}
\end{align}
$$

### Final Results:

- **Slope**: $\dfrac{2}{5}$
- **$y$-intercept**: $\left(0, -\dfrac{3}{5}\right)$
- **$x$-intercept**: $\left(\dfrac{3}{2}, 0\right)$

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## 18. Determine the slope and intercepts of the line or state that they do not exist: $x=1$.

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<summary> Solution: </summary>

Given the equation of the line:

$$
x = 1
$$

### Step 1: Interpret the equation

The equation $x = 1$ represents a **vertical line** passing through $x = 1$. Vertical lines have the following properties:

- **Slope**: The slope of a vertical line is **undefined** because the change in $x$ is zero, and division by zero is undefined.

- **$y$-intercept**: A vertical line does not have a $y$-intercept unless it coincides with the $y$-axis, which in this case it does not.

- **$x$-intercept**: The $x$-intercept is where the line crosses the $x$-axis. For $x = 1$, the $x$-intercept is at $(1, 0)$.

### Final Results:

- **Slope**: Undefined
- **$y$-intercept**: Does not exist
- **$x$-intercept**: $(1, 0)$


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## Formulas

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

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

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