PLZZ HELP!! I just dont understand it correctly
1 answer:
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Isolate the root expression:
![\sqrt[3]{x+1}+2=0\implies\sqrt[3]{x+1}=-2](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%2B1%7D%2B2%3D0%5Cimplies%5Csqrt%5B3%5D%7Bx%2B1%7D%3D-2)
Take the third power of both sides:
![\sqrt[3]{x+1}=-2\implies(\sqrt[3]{x+1})^3=(-2)^3](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%2B1%7D%3D-2%5Cimplies%28%5Csqrt%5B3%5D%7Bx%2B1%7D%29%5E3%3D%28-2%29%5E3)
Simplify:
![(\sqrt[3]{x+1})^3=(-2)^3\implies x+1=-8](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7Bx%2B1%7D%29%5E3%3D%28-2%29%5E3%5Cimplies%20x%2B1%3D-8)
Isolate and solve for
:
Since the cube root function is bijective, we know this won't be an extraneous solution, but it doesn't hurt to verify that this is correct. When, we have
![\sqrt[3]{-9+1}=\sqrt[3]{-8}=\sqrt[3]{(-2)^3}=-2](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B-9%2B1%7D%3D%5Csqrt%5B3%5D%7B-8%7D%3D%5Csqrt%5B3%5D%7B%28-2%29%5E3%7D%3D-2)
as required.
Take the third power of both sides:
Simplify:
Isolate and solve for
Since the cube root function is bijective, we know this won't be an extraneous solution, but it doesn't hurt to verify that this is correct. When
as required.
Answer:
See below
Step-by-step explanation:
For the model function ,
is the vertex, and
is the horizontal growth or shrink factor.
We know that our vertex is from our graph, so we'll need to determine the value of
by using one of the points on the graph:
Thus, our equation for the given graph is .
The domain of a function is the set of all existing real x-values, thus, our domain for the given function is , or all real x-values.
The range of a function is the set of all existing real y-values, thus, our range for the given function is , or all real y-values starting with y=-1 which is included.