Question 116019
a)


From the graph of {{{f(x)=log(10,(x))}}} , we can see that only positive x values will work. In other words, the domain of f is  (0,*[Tex \LARGE \infty]). Now let's find the domain of {{{g(x)=log(10,(x+2))}}}:



{{{x+2>0}}} Set the inner expression greater than zero


{{{x>0-2}}} Subtract 2 from both sides



{{{x>-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,*[Tex \LARGE \infty])


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(x))}}} shifts the entire graph of   {{{f(x)=log(10,(x))}}}  two units to the left



Notice if we graph {{{f(x)}}} and {{{g(x)}}}, we get


{{{ graph( 500, 500, -10, 10, -10, 10, log(10,(x)),log(10,(x+2)))}}}  Graph of {{{f(x)=log(10,(x))}}} (red) and  {{{g(x)=log(10,(x+2))}}} (green)


and we can visually verify the transformation



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<b>Vertical Asymptote</b>:


From the graph, we can see that the vertical asymptote is {{{x=0}}} for {{{f(x)}}}. 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(x)}}} is {{{x=-2}}}


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



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<b>x-intercept in (x, y) form</b>:


From the graph, we can see that the x-intercept of {{{f(x)}}} 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(x)}}} is (-1,0)


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