Question

question
graph the function $f(x)=\log_{\frac{1}{4}}(x)$.
you can only graph integer points, for example you cannot plot the point $\left(-1,\frac{1}{4}\
ight)$. there is one point that can be moved above and below the $x$ axis as needed. do not move either point off of the graph as you will not be able to click on the point again.
provide your answer below:

Explanation:

Step1: Find key points for the logarithmic function

For a logarithmic function \( f(x)=\log_{a}(x) \), the key points are when \( x = 1 \) (since \( \log_{a}(1)=0 \) for any \( a>0,a
eq1 \)) and when \( x=a \) (since \( \log_{a}(a) = 1 \)) or \( x=\frac{1}{a} \) (since \( \log_{a}(\frac{1}{a})=- 1 \)).

For the function \( f(x)=\log_{\frac{1}{4}}(x) \):

• When \( x = 1 \): \( f(1)=\log_{\frac{1}{4}}(1)=0 \) (because any non - zero number to the power of 0 is 1, so \( (\frac{1}{4})^0 = 1 \)).
• When \( x=\frac{1}{4} \): \( f(\frac{1}{4})=\log_{\frac{1}{4}}(\frac{1}{4}) = 1 \) (because \( (\frac{1}{4})^1=\frac{1}{4} \)).
• When \( x = 4 \): \( f(4)=\log_{\frac{1}{4}}(4)=\log_{\frac{1}{4}}((\frac{1}{4})^{-1})=- 1 \) (using the property \( \log_{a}(a^{b})=b \) and \( 4 = (\frac{1}{4})^{-1} \)).

Since we can only plot integer points, the points with integer \( x \)-coordinates are \( (1,0) \) and \( (4, - 1) \), \( (\frac{1}{4},1) \) has a non - integer \( x \)-coordinate so we focus on \( (1,0) \) and \( (4,-1) \), \( (1,0) \) is already on the graph (the point on the x - axis at \( x = 1 \)), and we can plot \( (4,-1) \) and \( (\frac{1}{4},1) \) is not an integer \( x \), but if we consider integer \( x \) values, another point is when \( x = 1 \), \( y = 0 \); when \( x=4 \), \( y=-1 \); when \( x=\frac{1}{4} \), \( y = 1 \) (but \( x=\frac{1}{4} \) is not integer, so we use \( x = 1,y = 0 \); \( x = 4,y=-1 \); \( x=\frac{1}{4} \) is not allowed as per integer point rule, so the main integer points are \( (1,0) \) and \( (4,-1) \), and also when \( x=\frac{1}{4} \) is non - integer, but if we take \( x = 1 \), \( y = 0 \); \( x = 4 \), \( y=-1 \); and when \( x=\frac{1}{4} \) is not integer, so the key integer points are \( (1,0) \) and \( (4,-1) \), and also \( ( \frac{1}{4},1) \) is not integer, so we can plot \( (4,-1) \) and confirm \( (1,0) \).

Step2: Plot the points on the graph

We know that the domain of \( y = \log_{\frac{1}{4}}(x) \) is \( x>0 \). The function is a decreasing function because the base \( a=\frac{1}{4}<1 \).

The point \( (1,0) \) is on the x - axis. For \( x = 4 \), we calculate \( f(4)=\log_{\frac{1}{4}}(4)=\frac{\ln(4)}{\ln(\frac{1}{4})}=\frac{\ln(4)}{-\ln(4)}=-1 \), so the point \( (4,-1) \) is an integer point. Also, when \( x=\frac{1}{4} \), \( f(\frac{1}{4}) = 1 \), but \( x=\frac{1}{4} \) is not an integer, so we can't plot it as per the rule. So we plot the points \( (1,0) \) (already there) and \( (4,-1) \), and also the point \( (\frac{1}{4},1) \) is not allowed, so the main integer points to plot are \( (1,0) \) and \( (4,-1) \), and we can also note that when \( x = 1 \), \( y = 0 \); when \( x = 4 \), \( y=-1 \); and the function is decreasing, so as \( x \) increases, \( y \) decreases.

Answer:

To graph \( f(x)=\log_{\frac{1}{4}}(x) \), plot the points \((1,0)\) (on the x - axis) and \((4, - 1)\) (below the x - axis at \( x = 4 \)) and \((\frac{1}{4},1)\) is not an integer \( x \) so we use the integer points \((1,0)\) and \((4,-1)\). The graph is a decreasing curve passing through \((1,0)\) and \((4,-1)\) (and approaching the y - axis as \( x
ightarrow0^{+} \) and approaching \( -\infty \) as \( x
ightarrow+\infty \) but we plot the integer points \((1,0)\) and \((4,-1)\)).