Question

section 4.3 - logarithmic functions
question 8 of 13 (1 point) | question attempt: 1 of unlimited

graph the function.

$y = \log_4 x$

plot four points and an asymptote. then click on the graph icon.

Explanation:

Step1: Find the vertical asymptote

For the function \(y = \log_a x\), the vertical asymptote is \(x = 0\). Here \(a = 4\), so the vertical asymptote is \(x=0\).

Step2: Calculate points using the logarithmic property \(\log_a a^k=k\)

• For \(x = 1=4^0\), by the property \(\log_4 4^0 = 0\), so we get the point \((1,0)\).
• For \(x = 4=4^1\), by the property \(\log_4 4^1 = 1\), so we get the point \((4,1)\).
• For \(x=\frac{1}{4}=4^{-1}\), by the property \(\log_4 4^{-1}=-1\), so we get the point \((\frac{1}{4},-1)\).
• For \(x = 16=4^2\), by the property \(\log_4 4^2 = 2\), so we get the point \((16,2)\).

Plot these four points \((\frac{1}{4},-1)\), \((1,0)\), \((4,1)\), \((16,2)\) and the vertical asymptote \(x = 0\) on the coordinate plane. Then connect the points (for \(x>0\)) to graph the function \(y=\log_4 x\).

Answer:

To graph \( y = \log_4 x \), we can use the following key points and the vertical asymptote:

• Vertical Asymptote: \( x = 0 \) (since the domain of \( y=\log_a x\) is \(x>0\))
• Points:
• When \(x = 1\), \(y=\log_4 1=0\) (point \((1,0)\))
• When \(x = 4\), \(y=\log_4 4 = 1\) (point \((4,1)\))
• When \(x=\frac{1}{4}\), \(y=\log_4\frac{1}{4}=- 1\) (point \((\frac{1}{4},-1)\))
• When \(x = 16\), \(y=\log_4 16 = 2\) (point \((16,2)\))