In Exercises 25-34, begin by graphing f(x) = 2^x. Then use transf...

Question

In Exercises 25-34, begin by graphing f(x) = 2^x. Then use transformations of this graph to graph the given function. Be sure to graph and give equations of the asymptotes. Use the graphs to determine each function's domain and range. If applicable, use a graphing utility to confirm your hand-drawn graphs. g(x) = 2^(x+1)

Accepted Answer

Hello. Today we're gonna be graphing G of X as a transformation of F of X. And we're going to find the domain and range of G of X. So what we are given is F of X is equal to five to the power of X. Now this is a function of an exponential function in standard form. And what this means is that we have an exponential function that has a horizontal ASM tote at Y is equal to zero and it passes through the 00. comma one. And the shape of this graph is going to be a graph that approaches from negative infinity, very close to the ascent. Tote passed through zero comma one and an increase exponentially. So this is going to be the rough shape of F of X. Now let's go ahead and identify the transformations. We need to go from F of X to G of X. G of X is given to us as five to the power of X plus three. And we're only going to need one transformation to go from F of X to G of X. And that revolves around the quantity X plus three. C X plus three is in the form of x minus H. And whenever you have a quantity in the form of X minus H, that's going to indicate a horizontal shift. Currently X minus three is not in the standard form of x minus H but we can rewrite it to give us x minus negative three. And this is going to be where H is equal to negative three. That means that we're going to have a horizontal shift of three units to the left. So now let's go ahead and apply this transformation to the function F. Of X. So we're going to draw an axis here. And since we're going to shift this graph three units to the left, let's go ahead and use zero comma one as a reference point. C. We're going to be moving this 10.0.3 units to the left. So that means that we're going to be minus ng three to the X value of zero comma one. And that's going to give us the point negative three comma one. So let's go ahead and graph our new point which is at negative three comma one. Now this type of transformation doesn't affect the horizontal as in total. So we're still going to have a horizontal at y is equal to zero. And the transformation doesn't affect the shape of the graph. So we're still going to have the same shape of the graph is just now it passes through the point negative three comma one. And this is going to be the graph of G of X. Now that we have the graph of G of X. Let's go ahead and identify the domain and the range. The domain focuses on the least possible X value to the highest possible X value. So we're going to be reading the graph going from left to right. But as you can see here this graph approaches from negative infinity in the X axis. And as it increases is going to continue infinitely along the X axis. So the domain is going to be from negative infinity to positive infinity. Now let's go ahead and identify the range. The range focuses on the lowest possible Y value to the highest possible Y value. But notice that we have a horizontal ASM tote of Y is equal to zero. That means that this graph is going to increase or get closer to y is equal to zero as decreases but never equal zero. So with that being said, the range is going to start at zero but not include that value. So we're going to assign zero a parentheses. And as this graph increases it's going to approach infinity along the Y axis. So the range is going to end in positive infinity because it's never going to stop increasing and that's gonna be the domain and range and the shape of G of X. Now we just need to go ahead and choose the option that accurately represents this transformation and that option is going to be be. So, the answer to this problem is going to be be Because the graph here shows the original graph that has an intersecting point at 0: and a horizontal shift three units to the left which gives us a new point of negative three comma one. So the answer to this problem is going to be be. So I hope this video helps you in understanding how to approach this problem, and I'll go ahead and see you all in the next video.

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