Graph the given function. g ( x ) = − log 3 ( x + 2 ) + 1 g\left(x\right)=-\log_3\left(x+2\right)+1

In this problem, we're asked to graph the given function. And here, the function we're given is g of x is equal to log base 2 of x minus 1 minus 4. So let's go ahead and get started with step 0 and identify and graph our parent function because we're going to be dealing with a transformation here. So here at my function g of x, I see this log base 2 of x. So that tells me that my parent function, f of x, is equal to a log base 2 of x. Now that we know what our parent function is, we can go ahead and identify what points we need to plot and get those onto our graph. So my first point is going to be 1 over b negative one. Now here b is this 2, that's the base of my log, so this point is going to be onetwo-one. Then we always have that same point 10. And then finally, we have our point b 1. Now we just said b was 2, so there's my final point. Let's go ahead and get these plotted on our graph starting with 1 half negative one, and then we have 10, and finally 21. Now you can always plug in some more points if you want a bigger picture of what's going on here. But this is all we need in order to graph our new transformed graph. So let's go ahead and connect these points. And then we also want to go ahead and plot our vertical asymptote at x equals 0 using a dashed line right along that y axis. Now that we've plotted our parent function, we can go ahead and transform it in order to get our new function. So our first step in doing this is going to be to go ahead and shift that vertical asymptote that we just plotted to x equals h. Now looking at my function at g of x, I see this minus 1 here. And remember, whenever we're dealing with transformed functions, this will always look like x minus h. So here, my h value is just 1, positive 1. Then I can go ahead and plot that using a dashed line on my graph for my vertical asymptote. Now we can go ahead and move on to step 2. The first thing we wanna check for is if there is a reflection happening. Now remember, there's a reflection if we have a negative inside or outside of our funk of our new function, and we don't have that here. So we can go ahead and mark that out and move on to the second part of step 2, which is to shift our test points by h k. Now we already know our value for h is 1, but let's go ahead and identify our k value by looking back to our function g of x. Now we have that x minus h, but remember, it's always plus k on the end here. So my k value is this full negative 4, and we can go ahead and shift those test points that we originally plotted for our parent function in order to get the points of our new function. So we're going to shift them over 1 and down 4 based on these values of h and k. So starting with this first point here, I'm going to shift that positive one this way and then down 1, 2, 3, 4 in order to be right here. Then my next point as well, 1 over and a 1, 2, 3, 4 down. And then my last point, 1 over and 1, 2, 3, 4 down. Now that I have my new points, I can go ahead and connect these points approaching my asymptote. So here, it's gonna go up this way and down approaching that vertical asymptote as it's going the other direction. Direction. Now, of course, we want to go ahead and identify our domain and our range as well. So starting with our domain, we want to look at which direction our graph is approaching our asymptote from, which it's approaching it from the right here. It's coming in from the right side. So our domain is going to go from h, our asymptote value, to infinity. Now our asymptote is is at x equals 1, so our domain is going to go from h from 1 to infinity, and that's our domain. Now if it was coming from the left, it would be different. It would go from negative infinity until we reach that asymptote, but we don't have to worry about that here. Now for our range, our range is always going to be the same for all log functions. It's always going to be all real numbers, and that remains true here. So we can see that our original function just got shifted down and over a little bit. And we have our new transformed function g of x. Let me know if you have any questions.

6. Exponential & Logarithmic Functions

Graphing Logarithmic Functions

College Algebra

6. Exponential & Logarithmic Functions

Graphing Logarithmic Functions

Struggling with College Algebra?

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Multiple Choice

Graph the given function.
$g (x) = - \log_{3} (x + 2) + 1$

Graph of the function g(x) = -log₃(x+2) + 1, showing a blue curve and a vertical dashed line at x = -2.

Graph of the function g(x) = -log₃(x+2) + 1, showing a blue curve and a dashed vertical asymptote at x = -2.

Graph of the function g(x) = -log₃(x+2) + 1, showing a blue curve approaching a vertical asymptote at x = 1.

Graph of the function g(x) = -log₃(x+2) + 1, showing a curve with a vertical asymptote at x = -2.

Verified step by step guidance

Identify the base function: The given function is g(x) = \log_2(x-1) - 4. The base function is \log_2(x), which is a logarithmic function with base 2.

Determine the transformations: The function \log_2(x-1) represents a horizontal shift to the right by 1 unit. The '-4' at the end of the function indicates a vertical shift downward by 4 units.

Find the vertical asymptote: For the function \log_2(x-1), the vertical asymptote is at x = 1, because the logarithm is undefined for x ≤ 1.

Plot key points: Choose values of x greater than 1 to find corresponding y-values. For example, if x = 2, g(x) = \log_2(2-1) - 4 = \log_2(1) - 4 = 0 - 4 = -4. Plot this point and a few others to get the shape of the graph.

Sketch the graph: Draw the curve starting from just above the x-axis at x = 1, moving to the right, and approaching the vertical asymptote at x = 1. The graph should be decreasing and shifted downwards by 4 units.

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Struggling with College Algebra?

Graph the given function.g(x)=−log3(x+2)+1g\left(x\right)=-\log_3\left(x+2\right)+1

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Graphing Logarithmic Functions practice set

Graph the given function.
$g (x) = - \log_{3} (x + 2) + 1$