Transformations – Example

Senin, 20 Januari 2020

Example 1 Using transformations sketch the graph of the following functions.

(a)    g (x) = x² + 3

(b)    f (x) = x − 5

Solution

The first thing to do here is graph the function without the constant which by this point should be fairly simple for you. Then shift accordingly.

(a)    g (x) = x² + 3

In this case we first need to graph x2 (the dotted line on the graph below) and then pick this up and shift it upwards by 3. Coordinate wise this will mean adding 3 onto all the y coordinates of points on x².

Here is the sketch for this one.

(b)    f (x) = √x − 5

Okay, in this case we’re going to be shifting the graph of √x (the dotted line on the graph below) down by 5. Again, from a coordinate standpoint this means that we subtract 5 from the y coordinates of points on √x .

Here is this graph.

Example 2 Using transformations sketch the graph of the following functions.

    (a)       h (x) = (x + 2)³

Solution

(a)       h (x) = (x + 2)³

Okay, with these we need to first identify the “base” function. That is the function that’s being shifted. In this case it looks like we are shifting f (x) = x3 . We can then see that,

h (x) = (x + 2)³ = f (x +2)

In this case c = 2 and so we’re going to shift the graph of f (x) = x³ (the dotted line on the graph below) and move it 2 units to the left. This will mean subtracting 2 from the x coordinates of all the points on f (x) = x³.

Here is the graph for this problem.

In this case it looks like the base function is √x and it also looks like c = −4 and so we will be shifting the graph of √x (the dotted line on the graph below) to the right by 4 units. In terms of coordinates this will mean that we’re going to add 4 onto the x coordinate of all the points on √x.

Here is the sketch for this function.

Example 3 Use transformation to sketch the graph of each of the following.

(a)     f (x) = (x − 2)² + 4

(b)     g (x) =│x + 3│ − 5

Solution

(a)         f (x) = (x − 2)² + 4

In this part it looks like the base function is x² and it looks like will be shift this to the right by 2 (since c = −2 ) and up by 4 (since k = 4 ). Here is the sketch of this function.

(b)        g (x) =│x + 3│ − 5

For this part we will be shifting │x│ to the left by 3 (since c = 3) and down 5 (since k = −5).

Here is the sketch of this function.

Example 4 Using transformation sketch the graph of each of the following.

    (a)      g (x) = −x²

Solution

(a)   Based on the placement of the minus sign (i.e. it’s outside the square and NOT inside the square, or (−x)² ) it looks like we will be reflecting x² about the x-axis. So, again, the means that all we do is change the sign on all the y coordinates.

Here is the sketch of this function.

(b)  Now with this one let’s first address the minus sign under the square root in more general terms. We know that we can’t take the square roots of negative numbers, however the presence of that minus sign doesn’t necessarily cause problems. We won’t be able to plug positive values of x into the function since that would give square roots of negative numbers. However

So, don’t get all worried about that minus sign.

Now, let’s address the reflection here. Since the minus sign is under the square root as opposed to in front of it we are doing a reflection about the y-axis. This means that we’ll need to change all the signs of points on √x .

Note as well that this syncs up with our discussion on this minus sign at the start of this part.

Here is the graph for this function.

Sumber

Labels: Mathematician

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