bruno102
  • bruno102
Graph this f(x)=√x+2-3. I actually have an answer but don't know if it is correct.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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bruno102
  • bruno102
|dw:1378232185885:dw| mine was similar to this but more accurate on a graph
anonymous
  • anonymous
not exactly... do you understand the shifts and such?
bruno102
  • bruno102
Yes

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bruno102
  • bruno102
What I don't understand is how to change the equation to see the shifts (a, h, k)
anonymous
  • anonymous
set whatever is inside the sqrt = 0 and solve. that will get you h. the k is whatever is outside is the k
bruno102
  • bruno102
so I should solve √x+2
anonymous
  • anonymous
set x+2 = 0 and solve for x. that will be your h.
bruno102
  • bruno102
how am I supposed to solve if it is 0 I don't understand.
anonymous
  • anonymous
subtract 2 from both sides to solve for x
bruno102
  • bruno102
so f(x)=0-5?
bruno102
  • bruno102
I mean f(x)= √0-5
anonymous
  • anonymous
\[f(x) = \sqrt{x+2} - 3 \text{, yes or no?}\]
bruno102
  • bruno102
yes
anonymous
  • anonymous
so inside the square root, what value of x makes the inside = 0?
bruno102
  • bruno102
-2
anonymous
  • anonymous
great! that is h. in terms of x, is -2 to the left or right of the origin?
bruno102
  • bruno102
left?
anonymous
  • anonymous
correctamundo! now, that negative 3 is a shift in the y direction. is that going to be up or down?
bruno102
  • bruno102
down I believe
anonymous
  • anonymous
that's right! so your graph will pellet 2 to the left and down 3.
anonymous
  • anonymous
shift not pellet
bruno102
  • bruno102
How about the stretch and and reflection? (b and a)?
anonymous
  • anonymous
there isn't any of those in this problem. if f(x) is the parent, then g(x) = af(x-h) + k is a horizontal shift of h, a vertical shift of k and a stretch of a if |a|>1 or a compression if |a| <1. if a< 0, then it is a reflection.
bruno102
  • bruno102
see that is not what my teacher taught us, she said we have to factor the x+2 out of the equation to give us all of them but thanks.
anonymous
  • anonymous
i'm not sure what your teacher is going for but i'm really not seeing the connection here. good luck

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