Question

How would you describe the relationship between the real zeros and x-intercepts of the function mc033-1. Jpg? when you set the function equal to zero, the solution is x = 6; therefore, the graph has an x-intercept at x = 6. 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 = 3

Answer 1

When you set the function equal to zero, the solution is x = 3 and the graph has an x-intercept at x = 3.

Given the following data:

• $y=log_4(x-2)$

What is a logarithm function?

A logarithm function can be defined as a function that represents the inverse of an exponential function. Mathematically, a logarithm function is written as follows:

$y=log_ax$

The domain of the given logarithm function $y=log_4(x-2)$ is (0, infinity) and its graph wouldn't touch the vertical axis but moves to the right. Also, a real zero on the graph would only occurs at x = 1 because $log_4 1$ is equal to zero (0).

Similarly, the graph of $y=log_4(x-2)$ is equal to $log_4 x$ but it moves 2 units to the right.

In conclusion, the graph has an x-intercept at x = 3 when the function is set to zero (0) and its solution would be x = 3.

Read more on logarithm function here: brainly.com/question/26788007