In Exercises 53-58, begin by graphing f(x) = log₂ x. Then use tra...

Question

In Exercises 53-58, begin by graphing f(x) = log₂ x. Then use transformations of this graph to graph the given function. What is the vertical asymptote? Use the graphs to determine each function's domain and range. g(x) = log₂ (x + 1)

Accepted Answer

Hey, everyone. Here we are asked to graph the following function by graphing first, its parent function F of X is equal to log of X and performing transformation then determine its vertical Asymptote domain and range for G of X is equal to log of X plus three. Here we have four answer choice options, ABC and D each with some vertical Asymptote domain and range as well as a graph for G of X F of X and B vertical Asymptote. So beginning to solve this problem, we first want to graph the parent function given as F of X is equal to log of X. So to graph this function, we need to consider values of X. So we can first consider when X is equal to one. Well, when X is equal to one, we just need to find F of one which is equal to log of one and the log of one is just going to be zero. And so now we just want to consider other values of X and make a table of values so that we can graph our function. So we can have some of our X values be zero. 00.51234 and five. And then input these X values into our function. We get the Y values negative 0.30, positive 0.30 point 480.6. And lastly 0.7. So now with this table of values, we can go ahead and plot these values on a graph. And so plotting, we see we have this assortment of points on our graph. And so the next step after plotting the points is to just connect all of the points as a line. And we see that we get our F of X function, this log function of X. And we can notice here that the log this function of F of X does not touch the Y axis although it approaches the negative Y axis. And so now our second function that we need to graph is G of X is equal to log of X plus three. Well, while this function follows the form of G FX is equal to log base B of X plus C which is a transformation characterized by shifting the graph of F of X by left of C units. So that means we want to right or graph F G FX by just moving over F FX left by three units. So doing this, our graph becomes this graph where we now have F of X as the black line and then G of X three units to the left in red. And so it can be observed that the graph now has a vertical ascent tode here at X is equal to negative three because we see that G FX approaches negative three, X is equal to negative three, but never crosses over X is equal to negative three. So again here, our vertical sy tote is when X is equal to negative three. And then we see that the domain here, the domain is from negative three to positive infinity and negative three is not included because of our Asymptote. So it gets closed by a parentheses and then infinity is always excluded. So it also gets closed by a parentheses. And then our range looking at our Y A, we see we simply go from negative infinity to positive infinity. So again, our range is from negative infinity to positive infinity. And both of these infinities are closed by parentheses. So we see our final graph here includes our vertical ascent to, it also includes G of X as well as F of X. And so again, our vertical Asymptote is at X is equal to negative three. And we have identified our vertical Asymptote our domain and our range. So with all of these pieces of information, we can determine our final answer choice here, which is answer choice B where again, we have a vertical Asymptote at X is equal to negative three. A domain of from negative three excluded to positive infinity. And then a range from negative infinity to positive infinity. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

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In Exercises 53-58, begin by graphing f(x) = log₂ x. Then use transformations of this graph to graph the given function. What is the vertical asymptote? Use the graphs to determine each function's domain and range. g(x) = log₂ (x + 1)