In Exercises 95-106, begin by graphing the standard cubic functio...

Question

In Exercises 95-106, begin by graphing the standard cubic function, f(x) = x³. Then use transformations of this graph to graph the given function. h(x) = x³/2

Accepted Answer

Use the graph of the function F of X equals X to third minus one for applying transformation of graphs to graph the function according to the given condition G of X equals one third X to third minus one. So we want to see what happens when we go from F FX to G of X, we know less G of X just has one third multiplying the original F of X. This means we have a vertical stretch or shrink depending on our number. In this case, we have a vertical shrink of one thirds. This is because our multiplier one third is between zero and one. So we will use a vertical shrink. Let's find two values on our original function F of X and multiply both of them by one third to figure out what points we have on our new graph. Let's say you find their intercepts for instance, cent you wouldn't have Y equals zero, which means zero equals X to the third minus one. When it goes back to the third X is equals to one. So X intercept happens at 10. Why intercept? This is for F FX by the way, So I'll write that there. You have Y equals zero, the third minus one, which is just negative one. Give us a 10.0, negative one. Now our G FX points will be when we multiply our Ys by one third. So you have one third multiplied by the 10.10 for the wife that just gives us 10, one third multiplied by zero, negative one for the Y gives us zero, negative one third. We can now find the draft with two points, 100, negative one third. If you look at our possible graphs, we can determine answer B will have both of these points. OK? I hope to help you solve the problem. Thank you for watching. Goodbye.

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In Exercises 95-106, begin by graphing the standard cubic function, f(x) = x³. Then use transformations of this graph to graph the given function. h(x) = x³/2