graph the following function: $f(x)=\begin{cases}-x^{2}+2 &\text{if }x < 2\\3x - 4&\text{if }xgeq2end{cases}$

Explanation:

Step1: Analyze $y = -x^{2}+2$ for $x < 2$

This is a parabola. The general form of a parabola is $y = ax^{2}+bx + c$, here $a=-1$, $b = 0$, $c = 2$. The vertex of the parabola $y=-x^{2}+2$ is at $(0,2)$ (using the formula $x=-\frac{b}{2a}=0$). When $x = 2$, $y=-2^{2}+2=-2$. But since $x<2$, we have an open - circle at the point $(2, - 2)$ on this part of the graph.

Step2: Analyze $y = 3x - 4$ for $x\geq2$

This is a linear function. When $x = 2$, $y=3\times2-4=2$. So the graph of $y = 3x - 4$ starts at the point $(2,2)$ (a closed - circle since $x\geq2$). The slope of the line $y = 3x - 4$ is $m = 3$ and the $y$ - intercept is $b=-4$.

Answer:

To graph the function, first draw the parabola $y=-x^{2}+2$ for $x < 2$ with an open - circle at $(2,-2)$. Then draw the line $y = 3x - 4$ starting at the point $(2,2)$ (closed - circle) for $x\geq2$.