## High School Algebra I Unlocked (2016)

### Chapter 9. Manipulating Functions

### Lesson 9.3. Reflections

In __Chapter 7__, we discussed that the value of *a* in a quadratic function determines whether the parabola opens upwards or opens downwards. In the function *ƒ*(*x*) = *ax*^{2} + *bx* + *c*, the parabola will open upward if *a* is positive. If *a* is negative, the parabola will be reflected across the *x*-a*x*is and open downward. A **reflection** is a graph that results from flipping the original graph over a line.

The function *ƒ*(*x*) = *x*^{2} reflected across the *x*-axis as −*ƒ*(*x*) results in the following graphs:

Now, here’s where it gets a little crazy. If we were to then reflect the graph of −*ƒ*(*x*) across the *x*-axis, the resulting graph would be the original graph of *ƒ*(*x*); a reflection of a reflection is the original graph.

You can also reflect an image across the *y*-a*x*is, negating each instance of *x* in the function equation. Given the function *ƒ*(*x*) = *x*^{2}, the image of *ƒ*(−*x*) would be identical to the pre-image when reflected over the *y*-axis, due to the symmetrical nature of the parabola and its location on the *y*-axis.

The fact that *ƒ*(*x*) =*ƒ*(−*x*) is no small thing. In fact, you may encounter questions that ask you to determine whether a function is **even** or **odd**. But what does that mean? Whenever you have a function where *ƒ*(*x*) = *ƒ*(−*x*), the function is considered **even**; this means that the function is symmetrical about the *y*-axis. If you find that *ƒ*(−*x*) = −*ƒ*(*x*), the function is considered **odd**. And if neither *ƒ*(*x*) = *ƒ*(−*x*) nor*ƒ*(−*x*) = −*ƒ*(*x*), the function is considered neither even or odd.

For any function *ƒ*(*x*), if *ƒ*(*x*) = *ƒ*(−*x*), the function is **even.**

For any function *ƒ*(*x*), if *ƒ*(−*x*) = −*ƒ*(*x*), the function is **odd**.

For any function *ƒ*(*x*), if *ƒ*(*x*) ≠ *ƒ*(−*x*) and *ƒ*(−*x*) ≠ −*ƒ*(*x*), the function is neither even or odd.

**EXAMPLE **

**Determine algebraically whether the function v(x) = (x − 2)^{2} is even, odd, or neither.**

In order to determine if a function is even, odd, or neither, plug in −*x* to the function *v*(*x*) = (*x* − 2)^{2}, and then simplify as follows:

*v*(*x*) = (*x* − 2)^{2}

*v*(−*x*) = (−*x* − 2)^{2}

*v*(−*x*) = (−*x* − 2) (−*x* − 2)

*v*(−*x*) = *x*^{2} + 2*x* + 2*x* + 4

*v*(−*x*) = *x*^{2} + 4*x* + 4

*v*(−*x*) = (*x* + 2)^{2}

In this scenario, *ƒ*(*x*) ≠ *ƒ*(−*x*) and *ƒ*(−*x*) ≠ −*ƒ*(*x*), so the function is neither even or odd.

**EXAMPLE **

**Determine algebraically whether the function z(x) = 2x^{2} − 3 is even, odd, or neither.**

Just like in the previous example, plug in −*x* to the function *z*(*x*) = 2*x*^{2} − 3 and simplify to determine if a function is even, odd, or neither.

*z*(*x*) = 2*x*^{2} − 3

*z*(−*x*) = 2(−*x*)^{2} − 3

*z*(−*x*) = 2*x*^{2} − 3

That’s all there is to it! Here, we found that *z*(−*x*) = *z*(*x*), indicating that the graph of *z*(*x*) is identical to the graph of *z*(−*x*). Thus, the function *z*(*x*) is even.

Are you struggling to

understand that, given the

function *ƒ*(*x*) = *x*^{2},*ƒ*(*x*) = *ƒ*(−*x*)? Draw the

graph of *ƒ*(*x*) = *x*^{2} onto a

piece of paper, and fold

the paper along the *y*-axis.

If you hold the paper up

to the light you should see

a single line…because of

symmetry! Alternatively,

if you plug in −*x* for *x* in

the function *ƒ*(*x*) = *x*^{2}, you

will find that *ƒ*(−*x*) = *x*^{2}.

Thus, since *ƒ*(*x*) = *ƒ*(−*x*),

the function is even.

Let’s now consider the function *h*(*x*) = *x*^{3} + 4 and its reflection over the *x*-axis, −*h*(*x*).

You can also reflect a function over the *y*-axis. Take a look at the function *h*(*x*) = *x*^{3} + 4 and its reflection over the *y*-axis, *h*(−*x*):

−*ƒ*(*x*) reflects *ƒ*(*x*) about the *x*-axis.

*ƒ*(−*x*) reflects *ƒ*(*x*) about the *y*-axis.

Now that we’ve covered function reflections, let’s try some questions.

**Leave the NegativeExes Outside**

Having some difficulty

remembering when to

reflect a function over

the *x*- or *y*-axis? Use

this association: After

a rough breakup, you

may push a negative ex

out of your life. And in

math, if the negative

sign is outside the

function’s argument,

the graph is reflected

about the *x*-axis.

**EXAMPLE **

**If p(x) = x^{2} − 3, is the graph p(−x) + 1 reflected about the x- or y-axis? Graph both p(x) and p(−x) + 1 in the coordinate plane.**

There are three parts to this question: (1) determining whether *p*(−*x*) + 1 is a reflection of *p*(*x*) about the *x*- or *y*-axis, (2) graphing *p*(*x*), and (3) graphing *p*(−*x*) + 1, a transformation that involves a reflection and shift of *p*(*x*). First figure out whether *p*(−*x*) + 1 is a reflection of *p*(*x*) across the *x*- or *y*-axis. When any function *ƒ*(*x*) is transformed as *ƒ*(−*x*), the graph is reflected about the *y*-axis.

Next, graph the original function, *p*(*x*) = *x*^{2} − 3. Note that *p*(*x*) is simply a transformation of the most basic quadratic, *p*(*x*) = *x*^{2}. Here, −3 is not associated with the *x*-term, so simply shift the vertex of the graph of *x*^{2} down three units from the origin. Refer to the graph below.

Finally, graph *p*(−*x*) + 1, a transformation of *p*(*x*) that involves a reflection across the *y*-axis and a shift. If you were asked to reflect *p*(*x*) as *p*(−*x*), the two reflections would be identical. However, you are asked to graph *p*(−*x*) + 1. Since the positive 1 is not associated with the *x*-term, you will shift the graph of *p*(−*x*) up one unit on the *y*-axis. The resulting graphs of *p*(*x*), shown as a solid line, and *p*(−*x*) + 1, shown as a dashed line, are as follows:

Let’s try one more question before we move on.

**EXAMPLE **

**The graph above shows the function j(x) = (x − 2)^{3} as a solid line and the function k(x) as a dashed line. Which of the following could be the equation for k(x) ?**

**A) k(x) = −(x − 2)^{3}**

**B) k(x) = −(x + 2)^{3}**

**C) k(x) = (−x + 2)^{3} + 4**

**D) k(x) = (−x + 2)^{3} − 4**

This question requires you to identify the equation of *k*(*x*), represented by the dashed line in the provided graph, given that *j*(*x*) = (*x* − 2)^{3}. Tackle each answer choice one at a time. In (A), *k*(*x*) = −(*x* − 2)^{3}, *k*(*x*) is equal to −*j*(*x*), which means that *k*(*x*) is the graph of *j*(*x*) reflected about the *y*-axis. (Remember that when any function *ƒ*(*x*) is transformed as *ƒ*(−*x*), the graph is reflected about the *y*-axis.) In the graph provided, however, the function *j*(*x*) is not reflected across the *y*-axis to produce *k*(*x*); eliminate (A). Similarly, (B) represents a reflection of *j*(*x*) across the *y*-axis and, therefore, can be eliminated.

Now, you are left with (C) and (D), which are functions that have both a reflection and a shift. In fact, the only difference between the functions is the shift up or down by four units. In the graph provided, the zero of function *k*(*x*), where *x* = 2, is four units higher than that of *j*(*x*), indicating an upward shift. Since (D), *k*(*x*) = (−*x* + 2)^{3} − 4, indicates that the graph is shifted down by four units, it can be eliminated. In turn, the function *k*(*x*) in the provided graph is represented by the equation *k*(*x*) = (−*x* + 2)^{3} + 4, so the answer is (C).

The following graphs show the functions in (A), (B), and (D), where *k*(*x*) is represented as a dashed line, and the reflection of *k*(*x*) is shown as a solid line:

**Here is how you may see translations and reflections on the SAT.**

The graph of *y* = *g*(*x*) is shown in the figure above. If *g*(*x*) = *ax*^{2} + *bx* + *c* for constants *a, b,* and *c*, and if *abc* ¹ 0, then which of the following must be true?

A) *ac* > 1

B) *c* > 1

C) *ac* > 0

D) *a* > 0