Question

algebra 2a semester online practice
complete this assessment to review what youve learned. it will not count toward your grade.
which function would change the domain of $p(x) = -\frac{3}{2}sqrt{-(x - 1)} - 4$ to ${x|x \leq 5}$? (1 point)
\\(\circ\\) $j(x) = -\frac{3}{2}sqrt{-(x + 6)} - 4$
\\(\circ\\) $w(x) = -\frac{3}{2}sqrt{(x - 5)} - 4$
\\(\circ\\) $k(x) = -\frac{3}{2}sqrt{-(x - 5)} - 4$
\\(\circ\\) $m(x) = -\frac{3}{2}sqrt{-(x + 5)} - 4$
check answer remaining attempts 3

Explanation:

Step1: Find domain of \( p(x) \)

For \( p(x) = -\frac{3}{2}\sqrt{-(x - 1)} - 4 \), the radicand must be non - negative: \( -(x - 1)\geq0\Rightarrow x - 1\leq0\Rightarrow x\leq1 \). The domain is \( \{x|x\leq1\} \). We need a function with domain \( \{x|x\leq5\} \).

Step2: Analyze the radicand condition for the new domain

We need \( -(x - a)\geq0\Rightarrow x - a\leq0\Rightarrow x\leq a \). We want \( a = 5 \), so the radicand should be \( -(x - 5) \).

Step3: Check each option

• Option \( j(x) \): Radicand is \( -(x + 6) \), so \( x+6\leq0\Rightarrow x\leq - 6 \), not \( x\leq5 \).
• Option \( w(x) \): Radicand is \( (x - 5) \), so \( x - 5\geq0\Rightarrow x\geq5 \), domain \( \{x|x\geq5\} \), not \( \{x|x\leq5\} \) (and also the radicand sign is wrong as original had a negative sign in front of the binomial inside the square root).
• Option \( k(x) \): Radicand is \( -(x - 5) \), so \( -(x - 5)\geq0\Rightarrow x - 5\leq0\Rightarrow x\leq5 \), which is the desired domain.
• Option \( m(x) \): Radicand is \( -(x + 5) \), so \( x + 5\leq0\Rightarrow x\leq - 5 \), not \( x\leq5 \).

Answer:

\( k(x)=-\frac{3}{2}\sqrt{-(x - 5)}-4 \) (the option corresponding to \( k(x) \))