Question

the graph shows function j, a transformation of $f(x)=x^{\frac{1}{2}}$
replace b and h to create the equation for function j
$j(x)=(bx + h)^{\frac{1}{2}}$

Explanation:

Step1: Recall the transformation formula

The general form of a horizontal transformation of \(y = x^{\frac{1}{2}}\) is \(y=(bx + h)^{\frac{1}{2}}\). For a horizontal shift \(y=(x - k)^{\frac{1}{2}}\) (when \(b = 1\)), the vertex of \(y=x^{\frac{1}{2}}\) is \((0,0)\) and for \(y=(x - k)^{\frac{1}{2}}\) the vertex is \((k,0)\).

Step2: Identify the vertex

The vertex of the function \(j(x)\) is \((4,0)\). Using the formula \(y=(x - k)^{\frac{1}{2}}\), when \(k = 4\), we have \(y=(x-4)^{\frac{1}{2}}\). Comparing with \(j(x)=(bx + h)^{\frac{1}{2}}\), we get \(b = 1\) and \(h=-4\).

Answer:

\(j(x)=(x - 4)^{\frac{1}{2}}\), so \(b = 1\) and \(h=-4\)