Question

how would you describe the relationship between the real zeros and x - intercepts of the function y = log₄(x - 2)?
when you set the function equal to zero, the solution is x = 6; therefore, the graph has an x - intercept at x =
when you substitute x = 0 into the function there is no solution; therefore, the graph will not have any x - intercepts.
since there is an asymptote at x = 2, the graph will not have an x - intercept; therefore, the function will have no real zeros.
when you set the function equal to zero, the solution is x = 3; therefore, the graph has an x - intercept at x =

Explanation:

Step1: Recall the definition of x - intercept

The x - intercept of a function \(y = f(x)\) is found by setting \(y = 0\). For the function \(y=\log_{4}(x - 2)\), when we set \(y = 0\), we have the equation \(0=\log_{4}(x - 2)\).

Step2: Use the property of logarithms

By the property of logarithms, if \(\log_{a}b=c\), then \(b=a^{c}\). For \(\log_{4}(x - 2)=0\), since \(a = 4\) and \(c = 0\), we know that \(x-2=4^{0}\).

Step3: Solve for x

Since \(4^{0}=1\), then \(x-2 = 1\), and \(x=3\). The x - intercept occurs when \(y = 0\), so the x - intercept is at the point \((x,0)\) where \(x = 3\).

Answer:

When you set the function equal to zero, the solution is \(x = 3\); therefore, the graph has an \(x\) - intercept at \(x = 3\).