SOLUTION: Could someon help me PLZ graph y=2(x+5)^-3 and label the vertex and all exact interceps, not to scale on the Y axis Thank you

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Question 58658: Could someon help me PLZ
graph y=2(x+5)^-3 and label the vertex and all exact interceps, not to scale on the Y axis
Thank you

Found 2 solutions by Edwin McCravy, funmath:
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
 Could someon help me PLZ 
graph y = 2(x+5)^-3 and label 
the vertex and all exact 
intercepts, not to scale on 
the Y axis 
Thank you

There is no "vertex".  There is no x-intercept.  
The y-intercept is

(0, 2/125).  Since the equation is equivalent to

        2 
y = ----------
     (x + 5)³

The denominator becomes zero at x = -5, thus the 
vertical line whose equation is x = -5 is a 
vertical asymptote.

Since the degree of the denominator, 2, is of 
greater than the degree of the numerator, 0, the 
x-axis, whose equatrion is y = 0, is the 
horizontal asymptote.

Find some points and you get the graph below.  
The vertical line is the vertical asymptote. The 
y-intercept (0, 2/125) is so close to the 
origin that it hardly discernable.  The curve 
does not really coincide with the x-axis but is 
so close to it, there is no way to indicate
that it is not actually coinciding with it.



Edwin

Answer by funmath(2933)   (Show Source): You can put this solution on YOUR website!
graph y=2(x+5)^-3 and label the vertex and all exact interceps, not to scale on the Y axis
On the off chance that you meant to type y=2(x+5)^2-3, I'm going to give this a try.
is vertex form, where the vertex is (h,k)

Vertex: (h,k)=(-5,-3)
:
Y-intercepts happen when x=0, let x=0 and solve for y.




There is a y-intercept at (0,47)
:
x-intercepts happen when y=0, let y=0 and solve for x using the square root property:








Some teachers are happy here.


The x intercepts are at (-5-sqrt(6)/2,0) and (-5+sqrt(6)/2,0)

Happy Calculating!!!

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