B) Graph each of the following:
i) \( f(x)=\left\{\begin{array}{ll}-x^{2} & \text { for } x \geq 0 \\ 2 x^{2} & \text { for } x<0\end{array}\right. \)
ii) \( f(x)=\left\{\begin{array}{c}2 x+3 \text { if } x<0 \\ x^{2} \text { if } 0 \leq x<2 \\ 1 \text { if } x \geq 2\end{array}\right. \)
iii) \( f(x)=\left\{\begin{array}{c}2 \text { if } x>2 \\ 1 \text { if } 0
Ask by Crawford King. in Zambia
Feb 20,2025
Solución de inteligencia artificial de Upstudy
Responder
Solución
Respondido por UpStudy AI y revisado por un tutor profesional


Extra Insights
To graph these piecewise functions, it's essential to first understand how they behave in different segments of the domain. Each function has specified rules for \(x\) in certain intervals, which creates distinct curves and lines that you will plot accordingly. For \(f(x)=\begin{cases}-x^{2} & \text { for } x \geq 0 \\ 2 x^{2} & \text { for } x<0\end{cases}\), plot a downward-opening parabola starting from the origin in the positive \(x\) direction, while in the negative \(x\) direction, you will plot an upward-opening parabola that starts from the origin. Make sure to highlight that the two parabolas meet at the origin, giving a smooth transition between the two. For the function \(f(x)=\begin{cases}2 x + 3 & \text{ if } x < 0 \\ x^{2} & \text{ if } 0 \leq x < 2 \\ 1 & \text{ if } x \geq 2\end{cases}\), first sketch a line with a positive slope (2) that intersects the \(y\)-axis at 3 for negative \(x\) values. Next, at \(x = 0\), transition into a quadratic curve starting from the origin and peaking at \(x = 2\). Finally, from \(x = 2\) onward, draw a straight horizontal line at \(y = 1\). Keep in mind the continuity and smoothness of transitions at points where the function rules change!
