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Chapter 1, Section 1.3, Intelligent Tutoring Question 004 Shifting graph Up Or down Suppose f is a function and & > 0. Define functions g and h by g(x) = f(x...

Question

Chapter 1, Section 1.3, Intelligent Tutoring Question 004 Shifting graph Up Or down Suppose f is a function and & > 0. Define functions g and h by g(x) = f(x) + a and h(x) = f(x)Thenthe graph of 9 is obtained by shifting the graph of f up a units_ the graph of h Is obtained by shifting the graph of f down a units_Part 1Your answer is correctAssume that fis " the function defined on the interval [1, 2] by the formula f(x) Sx2 + 5. The graph of g Is obtained by shifting the graph of f

Chapter 1, Section 1.3, Intelligent Tutoring Question 004 Shifting graph Up Or down Suppose f is a function and & > 0. Define functions g and h by g(x) = f(x) + a and h(x) = f(x) Then the graph of 9 is obtained by shifting the graph of f up a units_ the graph of h Is obtained by shifting the graph of f down a units_ Part 1 Your answer is correct Assume that fis " the function defined on the interval [1, 2] by the formula f(x) Sx2 + 5. The graph of g Is obtained by shifting the graph of f down 2 units Write the formula for 9 g(x) Sx +3 edlt



Answers

The graph of the function $g$ is formed by applying the indicated sequence of transformations to the given function $f$. Find an equation for the function g. Check your work by graphing fand $g$ in a standard viewing window.
The graph of $f(x)=x^{3}$ is shifted five units to the right and four units up.

This is problem # 37. We are given the following equation and are tasked with applying the following two transformations to it. So for our first step we take are Horizontal movement. In this case we have left by four. This affects our input. So we have the cube roots of X plus or minus four. Since we have a left movement, we will be adding. For Next we have a vertical movement so we'll have some sign plus or -5. Since the vertical movement is down, we'll be subtracting five. And this gives us G Fx. We can put both of these and gizmos first here's our fx, which is the cube root of X. Taking our first transformation gives us of expose four And then we subtract 5 to get glx.

Okay, so in Christian 41 trying to come up with a function g such that it matches, um, squared of X shifted to to the right and three down. So let's write out our equation first. Um, so we know we're going to need two constants because we're shifting both right and down. Um, so we're gonna need one inside the elementary function with X, we're never going to the right, so we know we need to subtract, and it's two units, and then we know we're going down in the vertical translation, so we need to subtract again, and that's by three units. So when we go to graph, um, there's no stretching or shrinking, and there's no reflection. So it's just the normal graph of the square root of X. Just moved, Um, so over two and down, three, 23 and then will look something like that

Find the equation to reapply these transformations. So Find the first one up by five. That gives us next to the third. That's five playing second transformation going right by one. That's expense. Want the Third power plus five. This is our question that we're looking for general those transformations are applied.

To answer this question, we can simply look at the graph and see how it has changed from the original function. So for F A. Becks, if we look at the Vertex, we can see that it has moved positive two units to the right and positive three units up. So we rewrite F of X in terms of the original function. Why, then we know that for every unit moved right or left, it affects the X value. So X minus two and then for every unit moved up her down, we can just add it to the US. So why at X minus two just replaces every instance off X and original function. Why, with X minus two. So f Quebec's could be written as the absolute value of X minus two two plus three. This is that books and we can use a similar approach for G of X. When we look at the graph, we can see that it has been flipped upside down, so we know that it is negative. Then we can see that the Vertex has been moved negative two units to the left and negative one unit down so similarly weaken right Dfx In terms of why so you can say G of X is equal to negative. Why at X plus two minus one, then we put in but the function get the negative absolute value of exposed to minus. This is G of X.


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