PLEASE HELP WITH ALGEBRA QUESTION NEED ASAP!!The graph of a logarithmic has a vertical asymptote at x=4 , contains the point (e+4, 1) , and has an x-intercept of 5. Which function is graphed?Question options:f(x) =ln(x−4)f(x) =ln(x+4)f(x) =lnx−4f(x) =lnx+4

Answer:The function f(x) = ln(x - 4) is graphed the question optionsStep-by-step explanation:* Lets study the the information of the problem- The graph of a logarithmic has a vertical asymptote at x=4* That means the curve gets closer and closer to the vertical line x = 4   but does not cross it- It contains the point (e+4, 1)* That means if we substitute x = e + 4 in the equation the value  of y will be equal to 1- It has an x-intercept of 5* That means if we substitute y = 0 in the equation the value of x   will be equal to 5* Lets find the right answer∵ f(x) = ln(x - 4)* To find the equation of the asymptote let x - 4 = 0∵ x - 4 = 0∴ x = 4 ∴ f(x) has a vertical asymptote at x = 4* Lets check the point (e + 4 , 1) lies on the graph of the f(x)∵ x = e + 4∴ f(e+4) = ln(e + 4 - 4) = ln(e)∵ ln(e) = 1∴ The point (e+4 , 1) lies on the graph of the function f(x)* To find the x-intercept put y = 0∵ f(x) = 0∴ ln(x - 4) = 0* Change the logarithmic function to the exponential function- The base of the ln is e∴ e^0 = x - 4∵ e^0 = 1∴ x - 4 = 1 ⇒ add 4 to the both sides∴ x = 5* The function f(x) = ln(x - 4) is graphed the question options