Graph the functions in Exercises 37–56.

y = (...

Question

Graph the functions in Exercises 37–56.

y = (x + 2)³/² + 1

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says for the given function, draw the graph, and we're given the function Y is equal to the quantity of X + 5 in quantity rates of the three halves plus 2. Below the problem we're given an empty graph on which to plot our function. So we need to draw the graph of Y is equal to the quantity of X + 5 in quantity rates to the 3 halves plus 2, and we're going to draw this using transformations. So we're gonna start by looking at a basic function, which is going to be F of X is equal to X-rays to the 3 has power. And if we look at our function Y, we see the quantity of X + 5 will shift our X value to the left by 5 units, and the plus 2 will shift our Y value up by 2 when compared to our function F of X. So what we're going to do is calculate a couple points for F of X and then use those points to draw our function Y. So we'll start off with looking at X equal to 0, so we'll have F of 0, which is going to be equal to 0 raised to the 3 house power, which is equal to 0. The next point is going to be when X is equal to 1, so we'll calculate F of 1, which is going to be equal to 1 raised to the 3 house power, which is going to be equal to 1. The next point we'll look at is when X is equal to 4, so we'll have F of 4, which is going to be equal to 4, raised to the 3 has power, which is going to be equal to 8 when we evaluate that expression. The next point is going to be when X is equal to 9, you'll have F of 9, which is going to be equal to 9, raised to the 3/2 power. Which is going to be equal to 27. And then finally, we'll look at the case when X is equal to 16, so we'll have F of 16, which is equal to 16, raised to the 3 half power, which is equal to 64. So we're going to use these points in our transformation in order to plot our function Y. So, if we look at the first point for F of X, we have the point 0.0. So, we need to transform that point to the correct location for our function Y. So, we need to shift to the X value to the left by 5 units, and the Y value up by 2 units. So that means that we're going to have a point at -5.2. The next point that we need to look at is for function F of X, which is 1.1. And again, we'll need to transform this point, so shifting it to left by 5 units and up by 2 units, we're going to have the point of minus 4.3. The next point that we need to look at is 4.8 for a function F of X, and again, Using our transformation of shifting to the left by 5 and up by 2, we're going to have the point of -1, 10. So the next point, we'll have 9.27, and again, shifting that to the left by 5 units and up by 2 units, we're going to have the point of minus 4.29. And then the final point that we need to look at is 16.24, and again, applying our transformation of shifting to the left by 5 and up by 2, we're going to have the point of 11. 66. So now we need to do is connect these points with a smooth curve. And in doing so, We will have the answer for this problem. And so we see that our function here is starting at the point. Of -5, 2, and gradually increasing. So, the graph that we have drawn on the screen here is the answer for this problem. So, I hope that this explanation has been useful, and I'll see you in the next video.

Graph the functions in Exercises 37–56.

y = (x + 2)³/² + 1

Key Concepts

Function Transformation

Domain and Range

Graphing Techniques

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