## 17Calculus Precalculus - Graph Symmetry in Rectangular Coordinates

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On this page, we cover symmetry of graphs in rectangular form. This other page covers symmetry of graphs in polar form.

We explain three types of graph symmetry on this page, even (y-axis), odd (origin) and x-axis symmetry. You will also come across symmetry described as symmetry about the x-axis, y-axis and y=x. Most graphs have no symmetry and some graphs can have more than one kind of symmetry. Knowing symmetry can sometimes simplify calculations in calculus.

Even Function

There are several ways to describe even symmetry, listed below.

 even symmetry symmetric about the y-axis even function

Here are a couple of examples of even functions.

$$y=x^2$$

$$y=\cos(x)$$

Looking at the graphs, notice that in each case, the graph on the left side of the y-axis is a mirror image of the graph on the right side. You can also think of it like this. If the point (a,b) is on the graph, then (-a,b) is also on the graph of an even function.
Mathematically, you can show that a function, $$f(x)$$ is even as follows.

 1 Find $$f(-x)$$ 2 If $$f(-x) = f(x)$$, then the function is even.

Let's do an example. In the graph on the left, we have $$f(x)=x^2$$.
$$f(-x) = (-x)^2 = x^2 = f(x)$$
Since $$f(x)=f(-x)$$, the function is even.

Odd Function

An odd function parallels the even function case, except that an odd function is mirrored or reflected about the origin. You can think of origin symmetry as a function reflected about the y-axis and then about the x-axis (or in reverse). Or you can think of the graph as rotating 180 degrees to get the same graph. Using points, you can notice that for a point (a,b) on the graph, the point (-a,-b) will also be on the graph of an odd function.

 odd symmetry symmetric about the origin odd function

Here are a couple of examples.

$$y=x^3$$

$$y=\sin(x)$$

We follow the same procedure with the equations that we did with an even function, except in this case, for a function to be odd, $$f(x) = -f(-x)$$. Let's do an example.

$$g(x) = x^3$$
$$g(-x) = (-x)^3 =$$ $$-x^3 = -(x^3) = -g(x)$$
Since $$g(-x) = -g(x)$$, the function $$g(x)=x^3$$ is odd.

x-axis Symmetry

Here is a graph showing x-axis symmetry. Similar to an even function that is symmetric about the y-axis, this graph has a reflection across the x-axis. Notice that it is not a function since it does not pass the vertical line test. This is true of all graphs with x-axis symmetry. One way to think about x-axis symmetry is to notice if we have a point (a,b) on the graph, the point (a,-b) is also on the graph.

$$y^2=x$$

Before jumping into some practice problems, let's watch this short video to make sure you are clear about the three kinds of symmetry.

### MIP4U - 3 kinds of symmetry [6min-45secs]

video by MIP4U

Okay, time for some practice problems.

Practice

Unless otherwise instructed, determine the symmetry of these graphs of these equations using algebraic techniques.

$$f(x) = x^2 + 1$$

Problem Statement

Determine the symmetry of the graph of the equation $$f(x) = x^2 + 1$$ using algebraic techniques.

$$f(x)$$ has even symmetry

Problem Statement

Determine the symmetry of the graph of the equation $$f(x) = x^2 + 1$$ using algebraic techniques.

Solution

First, let's test for even symmetry.
$$\begin{array}{rcl} f(x) & = & x^2+1 \\ f(-x) & = & (-x)^2+1 \\ & = & x^2+1 \\ & = & f(x) \end{array}$$
Since $$f(-x) = f(x)$$, this function has even symmetry.
If we look more closely at the equation, this is a parabola with vertex at the origin shifted up one unit.
Although the problem did not state that we needed a graph, we have provided one here to check if our answer makes sense.

$$f(x)$$ has even symmetry

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$$y=-x^4+4x^2+5$$

Problem Statement

Unless otherwise instructed, determine the symmetry of the graph of the equation $$y=-x^4+4x^2+5$$ using algebraic techniques.

Solution

### 1497 video

video by MIP4U

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$$y=x^3-2x$$

Problem Statement

Unless otherwise instructed, determine the symmetry of the graph of the equation $$y=x^3-2x$$ using algebraic techniques.

Solution

### 1498 video

video by MIP4U

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$$x^2+4y^2=4$$

Problem Statement

Unless otherwise instructed, determine the symmetry of the graph of the equation $$x^2+4y^2=4$$ using algebraic techniques.

Solution

### 1499 video

video by MIP4U

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$$y^2-2y^4=6x$$

Problem Statement

Unless otherwise instructed, determine the symmetry of the graph of the equation $$y^2-2y^4=6x$$ using algebraic techniques.

Solution

### 1500 video

video by MIP4U

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$$f(x) = x^4 - 29x^2 + 100$$

Problem Statement

Determine the symmetry of the graph of the equation $$f(x) = x^4 - 29x^2 + 100$$ using algebraic techniques.

Solution

### 2551 video

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$$y = x(x-2)(x+3)$$

Problem Statement

Determine the symmetry of the graph of the equation $$y = x(x-2)(x+3)$$ using algebraic techniques.

Solution

### 2552 video

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$$f(x) = x^3 - 9x$$

Problem Statement

Determine the symmetry of the graph of the equation $$f(x) = x^3 - 9x$$ using algebraic techniques.

Solution

### 2553 video

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$$f(x) = 4x^2 - 2x$$

Problem Statement

Determine the symmetry of the graph of the equation $$f(x) = 4x^2 - 2x$$ using algebraic techniques.

Solution

### 2554 video

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$$f(x) = 2x^2 + x^4 + 1$$

Problem Statement

Determine the symmetry of the graph of the equation $$f(x) = 2x^2 + x^4 + 1$$ using algebraic techniques.

Solution

### 2555 video

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$$f(x) = x^3 - x$$

Problem Statement

Determine the symmetry of the graph of the equation $$f(x) = x^3 - x$$ using algebraic techniques.

Solution

### 2556 video

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Really UNDERSTAND Precalculus

 functions vertical line test

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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