SOLUTION: Can someone help me with this graphing? Describe the transformations on the following graph of f(x)=log(x). State the replacement of the verticle asymptote and x-intercept after

Algebra ->  Graphs -> SOLUTION: Can someone help me with this graphing? Describe the transformations on the following graph of f(x)=log(x). State the replacement of the verticle asymptote and x-intercept after      Log On


   



Question 116017: Can someone help me with this graphing?
Describe the transformations on the following graph of f(x)=log(x). State the replacement of the verticle asymptote and x-intercept after the transformation. For example, "left 1" or stretched vertically by a factor of 2" or descriptions.On my graph i see a line from -2 on the y axis and 10 on the x-axis
a)g(x)=log(x+2)
Description of transformation:
Vertical asymptote:
x-intercept in (x,y) form:
b) g(x)= -log(x)
Description of transformation:
Vertical asymptote:
x-intercept in (x,y)form:
Need help asap
Thankyou

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
a)

From the graph of f%28x%29=log%2810%2C%28x%29%29 , we can see that only positive x values will work. In other words, the domain of f is (0,). Now let's find the domain of g%28x%29=log%2810%2C%28x%2B2%29%29:


x%2B2%3E0 Set the inner expression greater than zero

x%3E0-2 Subtract 2 from both sides


x%3E-2 Combine like terms on the right side


So that means x must be greater than -2
So here is the domain in interval notation: (-2,)

Notice how the endpoint of the domain has been shifted to the left two units. So what this did was simply shift every x value 2 units to the left

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Answer:
So the transformation g%28x%29%29 shifts the entire graph of f%28x%29=log%2810%2C%28x%29%29 two units to the left


Notice if we graph f%28x%29 and g%28x%29, we get

Graph of f%28x%29=log%2810%2C%28x%29%29 (red) and g%28x%29=log%2810%2C%28x%2B2%29%29 (green)

and we can visually verify the transformation




Vertical Asymptote:

From the graph, we can see that the vertical asymptote is x=0 for f%28x%29. Since we've shifted the graph 2 units to the left, we've also shifted the vertical asymptote 2 units to the left.

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Answer:
So the vertical asymptote for g%28x%29 is x=-2

We can visually verify this if we look at the graph above




x-intercept in (x, y) form:

From the graph, we can see that the x-intercept of f%28x%29 is (1,0). Since we've shifted everything two units to the left, the x-intercept shifts two units to the left also.

So subtract 2 from 1 to get

(1-2,0)---->(-1,0)

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Answer:
So the x-intercept of g%28x%29 is (-1,0)

Once again, we can visually verify this if we look at the graph above








b)

Description of transformation:

Remember, f%28x%29 is the same as y. So this means y=log%28x%29


Now if we negate both sides to get -y=-log%28x%29

So g%28x%29 is simply making each y coordinate becomes it's opposite. So something like (0,2) becomes (0,-2) and (3,-2) becomes (3,2), etc


----------------------------------
Answer:
So what's happening is that the graph is being reflected over the x-axis


Notice if we graph f%28x%29 and g%28x%29, we get

Graph of f%28x%29=log%2810%2C%28x%29%29 (red) and g%28x%29=-log%2810%2C%28x%29%29 (green)

and we can visually verify the transformation





Vertical Asymptote:

From the graph, we can see that the vertical asymptote is x=0. Since the transformation reflected the graph across the x-axis, the vertical asymptote of g%28x%29 is the same as the vertical asymptote of f%28x%29

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Answer:

So the vertical asymptote of g%28x%29 is x=0


We can verify this by looking at the graph above



x-intercept in (x, y) form:

From the graph, we can see that the x-intercept of f%28x%29 is (1,0). Since we've reflected everything with respect to the x-axis, the point on the x-axis is not affected. In other words the x-intercept of g%28x%29 is the same as the x-intercept of f%28x%29

----------------------------------
Answer:
So the x-intercept of g%28x%29 is (1,0)


Once again, we can visually verify this if we look at the graph above