Sketching the Parabola Exercise 1: \( \begin{array}{lll}\text { 1) Sketch the graphs of the } f_{0} \\ \text { (a) } y=2(x-3)^{2}-2 & \text { c) } h(x)=-(x-3)^{2}+5 & \text { e) } m(x)=-(x+2)^{2}-1 \\ \text { /b) } f(x)=-(x+2)^{2}+4 & \text { d) } p(x)=1 / 2(x+2)^{2}+2 & \text { f) } f: x \rightarrow(x+3)^{2}-4\end{array} \)
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The Deep Dive
To sketch the parabolas, you’ll want to identify their vertex and direction of opening. For example, in \(y=2(x-3)^{2}-2\), the vertex is at (3, -2) and it opens upwards because the coefficient of the squared term is positive. Meanwhile, \(h(x)=-(x-3)^{2}+5\) has a vertex at (3, 5) and opens downwards, meaning it peaks at that vertex. Now, a common mistake when sketching these parabolas is overlooking the impact of transformations. Remember that the vertex form \(y=a(x-h)^{2}+k\) allows you to quickly identify the vertex at (h, k). Make sure to pay attention to the coefficient \(a\); if it's positive, the parabola opens up, and if negative, it opens down. Happy sketching!