Graph the rational function using transformations. f ( x ) = 1 ( x + 3 ) 2 − 2 f\left(x\right)=\frac{1}{\left(x+3\right)^2}-2 f ( x ) = ( x + 3 ) 2 1 − 2

In this problem, we're asked to graph the rational function using transformations, and here I have the function g of x is equal to 1 over x plus 3 squared minus 2. So I'll be transforming the function 1 over x squared. Now starting with step 1 here, we want to go ahead and plot our vertical asymptote at x equals h. And here I have x plus 3, so we need to be careful with the sign here. Remember, when we're dealing with transformations, it's always x minus h. So if I have a plus here, that tells me that this is x minus negative 3. That double negative gives me that positive. So this vertical asymptote will be at x equals negative 3, taking that sign into account. So plotting that on my graph here using a dash line, now I can move on to my horizontal asymptote at y equals k. Now y equals k is at negative 2 here, so I'm gonna go ahead and plot that using a dash line at y equals negative 2. Now moving on to step 3, we need to determine if there is a reflection and there is not a reflection. I do not see any extra negatives there, so I don't have to worry about that. But moving on to 3 b, we do need to go ahead and shift those test points by h comma k, in this case, by negative 3 and negative 2. So let's go ahead and do that. Now with my points here, I wanna move them 3 units to the left. So starting with this point, 3 units to the left and then 2 units down. So right here is gonna be my first point. And then my second point, same thing, 3 units to the left and then 2 units down. So 3 left, 1, 2, 3, and 2 down. So I'm going to end up right here. Now that we have shifted our points we can go ahead and move on to our final step which is going to be to sketch our curves approaching our asymptotes. So going back to our graph over here I can go ahead and sketch those curves. No reflection happens so we're going to have the exact same shape in the exact same domain will always go from negative infinity to the asymptote so in this case our domain will always go from negative infinity to that asymptote. So in this case, from negative infinity to negative 3, and then from negative 3 to positive infinity. And then finally our range here since we are only in that part above our vertical or above our horizontal asymptote this is going to go from my horizontal asymptote to infinity. So it goes up to infinity, but from my horizontal asymptote at negative 2, so from negative 2 to infinity. Now that we've identified that domain in our range we are completely done here, thanks for watching and I'll see you in the next one.

5. Rational Functions

Graphing Rational Functions

College Algebra

5. Rational Functions

Graphing Rational Functions

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Multiple Choice

Graph the rational function using transformations.
$f\left(x\right)=\frac{1}{\left(x+3\right)^2}-2$

Graph of f(x)=1/(x+3)^2-2 showing vertical and horizontal asymptotes.

Graph of f(x)=1/(x+3)^2-2 with vertical asymptote at x=-3.

Graph of f(x)=1/(x+3)^2-2 illustrating behavior near vertical asymptote.

Graph of f(x)=1/(x+3)^2-2 showing a horizontal asymptote at y=-2.

Verified step by step guidance

Identify the base function: The base function here is $ f(x) = \frac{1}{x^2} $, which is a rational function with a vertical asymptote at $ x = 0 $ and a horizontal asymptote at $ y = 0 $.

Apply horizontal shift: The function $ f(x) = \frac{1}{(x+3)^2} $ indicates a horizontal shift to the left by 3 units. This moves the vertical asymptote from $ x = 0 $ to $ x = -3 $.

Apply vertical shift: The function $ f(x) = \frac{1}{(x+3)^2} - 2 $ indicates a vertical shift downward by 2 units. This moves the horizontal asymptote from $ y = 0 $ to $ y = -2 $.

Analyze the transformations: The graph of $ f(x) = \frac{1}{(x+3)^2} - 2 $ will have a vertical asymptote at $ x = -3 $ and a horizontal asymptote at $ y = -2 $. The graph will approach these asymptotes but never touch them.

Sketch the graph: Start by drawing the asymptotes at $ x = -3 $ and $ y = -2 $. Then, sketch the curve of the function, which will be similar to $ \frac{1}{x^2} $ but shifted according to the transformations. The graph will be symmetric about the vertical asymptote.