Question

graph the piecewise function given below.
$f(x) = \

$$\begin{cases} 5 & \\text{for } x < 1 \\\\ (x - 4)^2 & \\text{for } 1 \\leq x \\leq 7 \\end{cases}$$

$

Explanation:

Step1: Analyze the first piece ($x < 1$)

The first part of the piecewise function is $f(x)=5$ for $x < 1$. This is a horizontal line. We plot a horizontal line at $y = 5$ for all $x$ values less than 1. At $x = 1$, since the domain for this piece is $x < 1$, we use an open circle at the point $(1,5)$ to indicate that $x = 1$ is not included in this part of the function.

Step2: Analyze the second piece ($1\leq x\leq7$)

The second part is $f(x)=(x - 4)^2$ for $1\leq x\leq7$. This is a quadratic function in vertex form $y=a(x - h)^2+k$, where the vertex is at $(h,k)=(4,0)$.

• First, find the value at $x = 1$: $f(1)=(1 - 4)^2=(- 3)^2 = 9$. So we have a closed circle at $(1,9)$ (since $x = 1$ is included in this domain).
• Find the value at $x = 4$: $f(4)=(4 - 4)^2=0$, so the vertex is at $(4,0)$.
• Find the value at $x = 7$: $f(7)=(7 - 4)^2=(3)^2 = 9$, so we have a closed circle at $(7,9)$ (since $x = 7$ is included in this domain).
• We can also find a few more points to sketch the parabola. For example, at $x=2$: $f(2)=(2 - 4)^2=(-2)^2 = 4$; at $x = 3$: $f(3)=(3 - 4)^2=(-1)^2 = 1$; at $x=5$: $f(5)=(5 - 4)^2=1$; at $x = 6$: $f(6)=(6 - 4)^2=4$. Then we sketch the parabola opening upwards (since the coefficient of $(x - 4)^2$ is positive) between $x = 1$ and $x = 7$, connecting the points $(1,9)$, $(2,4)$, $(3,1)$, $(4,0)$, $(5,1)$, $(6,4)$, $(7,9)$ with a smooth curve.

Step3: Combine the two pieces

• The first piece is a horizontal line $y = 5$ with an open circle at $(1,5)$.
• The second piece is a parabola from $x = 1$ (closed circle at $(1,9)$) to $x = 7$ (closed circle at $(7,9)$) with vertex at $(4,0)$.

Answer:

To graph the piecewise function:

1. For \(x < 1\), draw a horizontal line \(y = 5\) with an open circle at \((1,5)\).
2. For \(1\leq x\leq7\), draw the parabola \(y=(x - 4)^2\) with a closed circle at \((1,9)\) (since \(f(1)=(1 - 4)^2 = 9\)), vertex at \((4,0)\), and a closed circle at \((7,9)\) (since \(f(7)=(7 - 4)^2=9\)). The parabola is symmetric about \(x = 4\) and passes through points like \((2,4)\), \((3,1)\), \((5,1)\), \((6,4)\) within the domain \(1\leq x\leq7\).