Graphing Logarithmic Functions Practice

We can use several methods to graph $y=4\log_3(x+1)$, but let's use transformations.

The graph of $y=\log_3x$ can be transformed to get the graph of $y=4\log_3(x+1)$. 

Note it is important to follow order of operations when building the function, as function transformations are not always commutative. 

Replacing $x$ with $x+1$ shifts the graph of $y=\log_3x$ to the left $1$ unit.

Since the graph of $y=\log_3x$ has a vertical asymptote of $x=0$, the graph of $y=\log_3(x+1)$ has a vertical asymptote of $x=-1$.

Multiplying $\log_3(x+1)$ by $4$ stretches the graph of
$y=\log_3(x+1)$ vertically by a factor of $4$.

Note that the graph of the function passes through $(2,1)$ before the stretch, and $(2, 4)$ after the stretch. This is because each $y$-coordinate is multiplied by $4$.

The above graph has an intercept of $(0,0)$ and passes through $(2,4)$.

We can verify this algebraically to check our work. 

The correct graph is graph A:

The strategy

We need to determine the position of the asymptote, and find two points on the graph.

Determining the vertical asymptote

Let's find the vertical asymptote of the graph of $y=-4\log_2(x-5)$.

From the parent function $y=\log_2x$, we see that the vertical asymptote of a logarithmic function occurs when the argument is equal to $0$.

In this case, that's when $x-5=0$ or when $x=5$.

Determining the $x$-intercept

We can find the $x$-intercept of the graph by finding the $x$-value that makes $y=0$.

The $x$-intercept is $(6,0)$. 

Finding another point on the graph

We need yet another point on the graph of $y=-4\log_2(x-5)$.

When thinking about what

$x$-values to use as inputs, it is "nice" to choose $x$-values that result in the argument equating to a power of $2$ as it makes for easy calculations and graphing!

For example, ${x=7}$ makes an argument of $2$, and so we can find the output as follows:

$\begin{aligned}y&=-4\log_2(x-5)\\\\&=-4\log_2(7-5)\\\\&=-4\log_2(2)\\\\&=-4\cdot 1\\\\&=-4\end{aligned}$

The graph passes through $(7,-4)$.

In conclusion, the graph has a vertical asymptote at $x=5$, and it passes through $(6,0)$ and $(7,-4)$. Therefore, the correct graph looks as follows:

First, notice that the graph of $y=\log_2x$ has a vertical asymptote. The graphs of $A$ and $D$ have horizontal asymptotes, not vertical asymptotes, so right away we can eliminate those options.

We can use several methods to graph $y=-\log_2(x+4)$, but let's use transformations.

The graph of $y=\log_2x$ can be transformed to get the graph of $y=-\log_2(x+4)$. 

Note it is important to follow order of operations when building the function, as function transformations are not always commutative. 

Replacing $x$ with $x+4$ shifts the graph of $y=\log_2x$ to the left $4$ units.

Since the graph of $y=\log_2x$ has a vertical asymptote of $x=0$, the graph of $y=\log_2(x+4)$ has a vertical asymptote of $x=-4$.

Multiplying $\log_2(x+4)$ by $-1$ reflects the graph of
$y=\log_2(x+4)$ across the $x$-axis.

The $x$-intercept of the above graph is $(-3,0)$ and the $y$-intercept is $(0,-2)$.We can verify this algebraically to check our work. 

The correct graph is graph B:

The strategy

We need to determine the position of the asymptote, and find two points on the graph.

Determining the vertical asymptote

Let's find the vertical asymptote of the graph of $y=4\log_2(x+6)-8$.

From the parent function $y=\log_2x$, we see that the vertical asymptote of a logarithmic function occurs when the argument is equal to $0$.

In this case, that's when $x+6=0$ or when $x=-6$.

Determining the $x$-intercept

We can find the $x$-intercept of the graph by finding the $x$-value that makes $y=0$.

The $x$-intercept is $(-2,0)$.

Finding another point on the graph

We need yet another point on the graph of $y=4\log_2(x+6)-8$.

When thinking about what $x$-values to use as inputs, it is "nice" to choose $x$-values that result in the argument equating to a power of $2$ as it makes for easy calculations and graphing!

For example, ${x=2}$ makes an argument of $8$, and so we can find the output as follows:

$\begin{aligned}y&=4\log_2(x+6)-8\\\\&=4\log_2(2+6)-8\\\\&=4\log_2(8)-8\\\\&=4\cdot 3-8\\\\&=4\end{aligned}$

The graph passes through $(2,4)$.

In conclusion, the graph has a vertical asymptote at $x=-6$, and it passes through $(-2,0)$ and $(2,4)$. Therefore, the correct graph looks as follows: