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

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

Step-by-step explanation:

The given function is .

So, for real zeros and x-intercept, the y-coordinate must be equal to zero.

Then it becomes

⇒ x - 2 = 1 {Since log 1 = 0}

⇒ x = 3

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

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.

the arc length is found by finding the circumference of the full circle (2xpixradius) and then dividing by 4 to find just one quarter of the circle (90 degrees).