Answer:
The graph has a domain of all real numbers.
The graph has a y-intercept at
.
The graph has an x-intercept at
.
Step-by-step explanation:
Given: The graph is ![y=\sqrt[3]{x-1}+2](https://tex.z-dn.net/?f=y%3D%5Csqrt%5B3%5D%7Bx-1%7D%2B2)
The domain of a function is a set of input values for which the function is real and defined.
Thus, the graph has a domain of
.
To find the y-intercept: To find the y-intercept, substitute
in
.
![\begin{aligned}y &=\sqrt[3]{x-1}+2 \\&=\sqrt[3]{0-1}+2 \\&=-1+2 \\&=1\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Dy%20%26%3D%5Csqrt%5B3%5D%7Bx-1%7D%2B2%20%5C%5C%26%3D%5Csqrt%5B3%5D%7B0-1%7D%2B2%20%5C%5C%26%3D-1%2B2%20%5C%5C%26%3D1%5Cend%7Baligned%7D)
Thus, the y-intercept is 
To find the x-intercept: To find the x-intercept, substitute
in
.
![\begin{aligned}y &=\sqrt[3]{x-1}+2 \\0 &=\sqrt[3]{x-1}+2 \\-2 &=\sqrt[3]{x-1} \\(-2)^{3} &=(\sqrt[3]{x-1})^{3} \\-8 &=x-1 \\-7 &=x\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Dy%20%26%3D%5Csqrt%5B3%5D%7Bx-1%7D%2B2%20%5C%5C0%20%26%3D%5Csqrt%5B3%5D%7Bx-1%7D%2B2%20%5C%5C-2%20%26%3D%5Csqrt%5B3%5D%7Bx-1%7D%20%5C%5C%28-2%29%5E%7B3%7D%20%26%3D%28%5Csqrt%5B3%5D%7Bx-1%7D%29%5E%7B3%7D%20%5C%5C-8%20%26%3Dx-1%20%5C%5C-7%20%26%3Dx%5Cend%7Baligned%7D)
Thus, the x-intercept is 
Answer:
0 less-than-or-equal-to x less-than infinity
Step-by-step explanation:
The square root function, y=√x, is defined for non-negative numbers. Its domain is ...
0 ≤ x < ∞
_____
<em>Comment on the question</em>
This shows up again and again in domain problems, so is worth remembering. The value under the radical cannot be negative (but it can be zero).
<span>if point A( 2,2) is reflected across the line Y then the new position A' is (-2,2) and the distance AY = distance A'Y
if A is reflected across line R it is now at point B and the distance AR = distance BR
lets say the point A(2,2) was perpendicular to the line R at the point (1, 4) then when reflected the point A now at location B will have coordinates</span><span>when flipped over a line of reflection the lengths are still the same
the point to the line of reflection is the same length as the line of reflection to the reflected position
the distance from the original point to the reflected point is twice the distance from the original point to the line of reflection
cannot see your polygon.
here is an example
</span>