Totally confused and have 5 more problems like this. Can someone explain each step for these questions? Thanks!
Let f(x) = e^x–2 + 3.
(a) Describe in words how the graph of f can be obtained from the graph of y = e^x.
(b) What is the domain of f?
(c) What is the range of f?
(d) What is the y-intercept? State the approximation to 2 decimal places (i.e., the nearest hundredth).
(e) What is the horizontal asymptote?
The graph is exponential, with f(x) shifted upward three units, and to the right two units.
domain of f is all real values of f
range? 3 to inf
y intercept is when x is zero. e^-2 + 3
or about 1/e^2 + 3 =appx 1/6 + 3 + 3.15 check that in your calculator
Sure! I'd be happy to help you understand each step for these questions.
(a) To describe how the graph of f can be obtained from the graph of y = e^x, we need to consider the transformation applied to the parent function y = e^x. In this case, f(x) = e^x–2 + 3 means that the graph of f is obtained by shifting the graph of y = e^x two units to the right and three units up.
(b) The domain of a function represents all the possible x-values for which the function is defined. In the case of f(x) = e^x–2 + 3, the domain is all real numbers. This means that you can plug in any real number for x and get a valid output for f(x).
(c) The range of a function represents all the possible y-values that the function can take. In this case, since e^x is always positive for any real value of x, subtracting 2 doesn't change the fact that the function output will always be greater than or equal to 1. Adding 3 further shifts the outputs higher. Therefore, the range of f(x) is y ≥ 4.
(d) To find the y-intercept, we need to find the value of y when x = 0. Plugging in x = 0 into f(x) = e^x–2 + 3, we get f(0) = e^0–2 + 3 = e^0–2 + 3 = 1–2 + 3 = 2. Therefore, the y-intercept is approximately 2.
(e) The horizontal asymptote represents the behavior of the function as x approaches positive or negative infinity. In the case of f(x) = e^x–2 + 3, as x approaches positive infinity, the exponential term e^x will increase without bound, causing f(x) to also increase without bound. Therefore, there is no horizontal asymptote for f(x) = e^x–2 + 3.
I hope this helps! Let me know if you have any more questions.