Question

graph the piece - wise function given below.
f(x)=\begin{cases}x^{2}-5&\text{for } - 1leq x<3\\|x - 5|+1&\text{for }xgeq4end{cases}
step 1: select a function then move with blue dot (if necessary)

Explanation:

Step1: Analyze \(y = x^{2}-5\) for \(-1\leq x<3\)

This is a parabola \(y = x^{2}\) shifted down 5 units. When \(x=-1\), \(y=(-1)^{2}-5=-4\); when \(x = 3\), \(y=3^{2}-5 = 4\) (but \(x = 3\) is not included in this part of the domain, so we have an open - circle at \((3,4)\)). The vertex of the parabola \(y=x^{2}-5\) is at \((0, - 5)\).

Step2: Analyze \(y=|x - 5|+1\) for \(x\geq4\)

The absolute - value function \(y = |x - 5|\) is a V - shaped graph with the vertex at \((5,0)\), and \(y=|x - 5|+1\) is shifted up 1 unit. When \(x = 4\), \(y=|4 - 5|+1=2\).

Answer:

Graph the parabola \(y=x^{2}-5\) for \(-1\leq x<3\) (closed - circle at \((-1,-4)\) and open - circle at \((3,4)\)) and the absolute - value function \(y = |x - 5|+1\) for \(x\geq4\) (closed - circle at \((4,2)\)).