Answer
To sketch the graph of \( y = \log_{2}(2x + 4) \), plot the x-intercept at \( \left(-\frac{3}{2}, 0\right) \) and the y-intercept at \( (0, 2) \). Draw a vertical asymptote at \( x = -2 \). The graph approaches the asymptote as \( x \) approaches \(-2\) from the right and increases without bound as \( x \) increases.
Solution
Function by following steps:
- step0: Find the \(x\)-intercept/zero:
\(y=\log_{2}{\left(2x+4\right)}\)
- step1: Set \(y\)=0\(:\)
\(0=\log_{2}{\left(2x+4\right)}\)
- step2: Swap the sides:
\(\log_{2}{\left(2x+4\right)}=0\)
- step3: Find the domain:
\(\log_{2}{\left(2x+4\right)}=0,x>-2\)
- step4: Convert the logarithm into exponential form:
\(2x+4=2^{0}\)
- step5: Evaluate the power:
\(2x+4=1\)
- step6: Move the constant to the right side:
\(2x=1-4\)
- step7: Subtract the numbers:
\(2x=-3\)
- step8: Divide both sides:
\(\frac{2x}{2}=\frac{-3}{2}\)
- step9: Divide the numbers:
\(x=-\frac{3}{2}\)
- step10: Check if the solution is in the defined range:
\(x=-\frac{3}{2},x>-2\)
- step11: Find the intersection:
\(x=-\frac{3}{2}\)
Analyze the y intercept of the function \( y=\log _{2}(2 x+4) \)
Function by following steps:
- step0: Find the y-intercept:
\(y=\log_{2}{\left(2x+4\right)}\)
- step1: Set \(x\)=0\(:\)
\(y=\log_{2}{\left(2\times 0+4\right)}\)
- step2: Multiply:
\(y=\log_{2}{\left(0+4\right)}\)
- step3: Simplify:
\(y=2\)
Find the vertical asymptotes of \( y=\log _{2}(2 x+4) \).
Function by following steps:
- step0: Find the vertical asymptotes:
\(y=\log_{2}{\left(2x+4\right)}\)
- step1: Evaluate the limit:
\(\lim _{x\rightarrow -2^{+}}\left(\log_{2}{\left(2x+4\right)}\right)\)
- step2: Rewrite the expression:
\(\log_{2}{\left(\lim _{x\rightarrow -2^{+}}\left(2x+4\right)\right)}\)
- step3: Calculate:
\(\log_{2}{\left(0\right)}\)
- step4: Calculate:
\(-\infty\)
- step5: \(x=-2\) is a vertical asymptote\(:\)
\(\begin{align}&x=-2\textrm{ }\textrm{is a vertical asymptote}\end{align}\)
- step6: List all vertical asymptotes of the function:
\(\begin{align}&x=-2\end{align}\)
To sketch the graph of the function \( y=\log _{2}(2x+4) \), we need to identify the intercepts and asymptotes.
### Step 1: Find the x-intercept
The x-intercept occurs when \( y = 0 \):
\[
0 = \log_{2}(2x + 4)
\]
This implies:
\[
2x + 4 = 1 \quad \Rightarrow \quad 2x = -3 \quad \Rightarrow \quad x = -\frac{3}{2}
\]
Thus, the x-intercept is at the point \( \left(-\frac{3}{2}, 0\right) \).
### Step 2: Find the y-intercept
The y-intercept occurs when \( x = 0 \):
\[
y = \log_{2}(2(0) + 4) = \log_{2}(4) = 2
\]
Thus, the y-intercept is at the point \( (0, 2) \).
### Step 3: Find the vertical asymptote
The vertical asymptote occurs where the argument of the logarithm is zero:
\[
2x + 4 = 0 \quad \Rightarrow \quad 2x = -4 \quad \Rightarrow \quad x = -2
\]
Thus, there is a vertical asymptote at \( x = -2 \).
### Summary of Key Points
- **X-intercept**: \( \left(-\frac{3}{2}, 0\right) \)
- **Y-intercept**: \( (0, 2) \)
- **Vertical asymptote**: \( x = -2 \)
### Sketching the Graph
1. Plot the x-intercept at \( \left(-\frac{3}{2}, 0\right) \).
2. Plot the y-intercept at \( (0, 2) \).
3. Draw a vertical dashed line at \( x = -2 \) to represent the asymptote.
4. The graph approaches the asymptote as \( x \) approaches \(-2\) from the right and increases without bound as \( x \) increases.
The graph will look like this:
```
|
3| *
2| (0,2)
1| /
| /
| /
---|-----/----------------
| /
| /
| /
| /
|/
|-----------------------
-3 -2 -1 0 1 2 3
```
This sketch shows the intercepts and the asymptote clearly.
Answered by UpStudy AI and reviewed by a Professional Tutor
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