Question

which statements describe the graph of $y = \sqrt3{x - 1} + 2$? select three options.
the graph has a domain of all real numbers.
the graph has a range of $y \geq 1$.
as $x$ is increasing, $y$ is decreasing.
the graph has a $y$-intercept at $(0, 1)$.
the graph has an $x$-intercept at $(-7, 0)$.

Explanation:

Step1: Analyze Domain

The function is \( y=\sqrt[3]{x - 1}+2 \). The cube root function \( \sqrt[3]{u} \) has a domain of all real numbers (since we can take the cube root of any real number \( u \)). Here, \( u=x - 1 \), so \( x-1 \) can be any real number, meaning \( x \) can be any real number. So the domain is all real numbers. This statement is correct.

Step2: Analyze Range

The cube root function \( \sqrt[3]{u} \) has a range of all real numbers (since for any real number \( y \), we can find a \( u \) such that \( \sqrt[3]{u}=y \)). Then \( y=\sqrt[3]{x - 1}+2 \) will also have a range of all real numbers (because adding 2 to all real numbers still gives all real numbers). So the statement "range of \( y\geq1 \)" is incorrect.

Step3: Analyze Monotonicity

The derivative of \( y = \sqrt[3]{x - 1}+2=(x - 1)^{\frac{1}{3}}+2 \) is \( y'=\frac{1}{3}(x - 1)^{-\frac{2}{3}}=\frac{1}{3\sqrt[3]{(x - 1)^{2}}} \). Since the denominator \( 3\sqrt[3]{(x - 1)^{2}} \) is always positive (except at \( x = 1 \) where the derivative is undefined, but the function is continuous there), the derivative is non - negative everywhere. So as \( x \) increases, \( y \) increases (or is constant, but in this case, since the derivative is positive except at a single point), so the statement "As \( x \) is increasing, \( y \) is decreasing" is incorrect.

Step4: Find y - intercept

To find the y - intercept, set \( x = 0 \). Then \( y=\sqrt[3]{0 - 1}+2=\sqrt[3]{-1}+2=- 1 + 2=1 \). So the y - intercept is at \( (0,1) \). This statement is correct.

Step5: Find x - intercept

To find the x - intercept, set \( y = 0 \). Then \( 0=\sqrt[3]{x - 1}+2 \), so \( \sqrt[3]{x - 1}=-2 \). Cubing both sides, we get \( x - 1=(-2)^{3}=-8 \), so \( x=-8 + 1=-7 \). So the x - intercept is at \( (-7,0) \). This statement is correct.

Answer:

The graph has a domain of all real numbers.
The graph has a y - intercept at \( (0,1) \).
The graph has an x - intercept at \( (-7,0) \)