# MAT 101 Online Lecture Notes: 3.2 Graphing Polynomial Functions

Polynomial functions have the following important property:

Every polynomial function of degree $n$ has at most $n$ real zeros.”

This property is called the Fundamental Theorem of Algebra.

As an application of this property, we see that a polynomial function of degree $n$ can have at most $n$ $x$-intercepts and at most $(n-1)$ turning points (local maximum and local minimum values).

Example. The function $f(x)=x^4-7x^3+12x^2+4x-16$ has three turning points that are two local minimum values and one local maximum value.

Graphing a Polynomial Function $p(x)$

1. First find all zeros of $p(x)$
2. Considering the even or odd multiplicity of each factor of $p(x)$, we can see the graph is crossing or touching the $x$-axis at each zero.
3. Use the leading-term test to determine the end behavior.
4. Use the $y$-intercept.

Example. Consider $f(x)=x^4-7x^3+12x^2+4x-16$. It can be factored as $f(x)=(x+1)(x-2)^2(x-4)$. (At this moment you don’t have to worry about how we get the facotring. We will discuss this in Sections 3.3 and 3.4.) So we find three zeros $-1$, $2$ (with multiplicity 2), and $4$. We can tell that the graph crosses at $-1$ and $4$ and touches the graph without crossing at $2$. Since the degree is $4$, an even number the graph goes up when $x\to -\infty$ and $x\to\infty$. These findings are all featured in the above graph.

The Intermediate Value Theorem

Let $p(x)$ be a polynomial (with real coefficients). Suppose that $p(a)$ and $p(b)$ have different signs for two distinct numbers $a$ and $b$. Then the graph of $p(x)$ must cross the $x$-axis between $a$ and $b$, i.e. $p(x)$ must have a zero between $a$ and $b$. This property is called the Intermediate Value Theorem.

Example. (a) Use the Intermediate Value Theorem to determine if
$$f(x)=x^3+3x^2-9x-13$$
has a zero between $a=1$ and $b=2$.

Solution. All you have to do is to evaluate $f(x)$ at $x=1$ and $x=2$, i.e. calculate $f(1)$ and $f(2)$ and see if they are different.
\begin{align*}
f(1)&=(1)^3+3(1)^2-9(1)-13=-18,\\
f(2)&=(2)^3+3(2)^2-9(2)-13=-11.
\end{align*}
$f(1)$ and $f(2)$ have the same sign, so the Intermediate Value Theorem won’t tell if $f(x)$ has a zero between $1$ and $2$. The following graph shows that it actually does not.

(b) Does $f(x)$ has a zero between $a=-5$ and $b=-4$?

Solution. $f(-5)=-18$ and $f(-4)=7$. Since their signs are different, by the Intermediate Value Theorem, there must be a zero between $-5$ and $-4$. The following graph confirms it.

# MAT 101 Online Lecture Notes: 3.1 Polynomial Functions and Models

A polynomial is a function of the form
$$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0.$$ The number $n$ is called the degree of the polynomial $P(x)$. The term $a_nx^n$ is called the leading term and $a_n$ is called the leading coefficient. The number $a_0$ is called the constant term. $P(x)$ is called linear if $n=1$, quadratic if $n=2$, cubic if $n=3$, quartic if $n=4$, quintic if $n=5$, sextic if $n=6$, septic if $n=7$, and so on so forth. You don’t really have to worry about memorizing these jargons. Names are less important. However you need to remember at least what the degree is and what the leading coefficient is.

There is a pattern for the long term behavior of a polynomial, i.e. the behavior of a polynomial when $x\to\infty$ or $x\to -\infty$. The behavior can be characterized as follows.

• $n=\mbox{even}$ and $a_n>0$:

Example. $f(x)=3x^4-2x^3+3$

• $n=\mbox{even}$ and $a_n<0$:

Example. $f(x)=-x^6+x^5-4x^3$

• $n=\mbox{odd}$ and $a_n>0$:

Example. $f(x)=x^5+\frac{1}{4}x+1$

• $n=\mbox{odd}$ and $a_n<0$:

Example. $f(x)=-5x^3-x^2+4x+2$

Finding zeros of a polynomial $P(x)$

By factoring, solve the equation $P(x)=0$. The solutions are the zeros of $P(x)$.

Example. Find the zeros of $P(x)=x^3+2x^2-5x-6$.

Solution. \begin{align*}
P(x)&=x^3+2x^2-5x-6\\
&=(x^3+x^2)+(x^2-5x-6)\ \mbox{(grouping)}\\
&=x^2(x+1)+(x-6)(x+1)\\
&=(x+1)(x^2+x-6)\\
&=(x+1)(x+3)(x-2).
\end{align*}
Hence, $P(x)$ has zeros $x=-3,-1,2$.

How do we determine whether $x=a$ is a zero of a polynomial $P(x)$?

To only check whether $x=a$ is a zero of $P(x)$, you don’t really have to factor $P(x)$. This is what you need to know. If $P(a)=0$, then $x=a$ is a zero of the polynomial $P(x)$.

Example. Consider $P(x)=x^3+x^2-17x+15$. Determine whether each of numbers 2 and $-5$ is a zero of $P(x)$.

Solution. $P(2)=(2)^3+(2)^2-17(2)+15=-7$, so $x=2$ is not a zero. $P(-5)=(-5)^3+(-5)^2-17(-5)+15=0$, so $x=-5$ is a zero of $P(x)$.

Even and Odd Multiplicity

Even and odd multiplicity is an important property for sketching the graph of a polynomial function. Suppose that $k$ is the largest integer such that $(x-c)^k$ is a factor of $P(x)$. The number $k$ is called the multiplicity of the factor $x-c$.

1. If $k$ is odd, the graph of $P(x)$ crosses the $x$-axis at $(c,0)$.
2. If $k$ is even, then the graph of $P(x)$ is tangent to the $x$-axis, i.e. touches the $x$-axis without crossing at $(c,0)$.

Example. Consider $f(x)=x^2(x+3)^2(x-4)(x+1)^4$. The factors $x$ and $x+3$ have multiplicity 2 and the factor $x+1$ has multiplicity 4. Hence the graph of $f(x)$ touches the $x$-axis without crossing at $x=0$, $x=-3$ and $x=-1$. The factor $x-4$ has multiplicity 1, so the graph crosses the $x$-axis at $x=4$. This is also shown in the following figure.

# MAT 101 Online Lecture Notes: 2.6 Solving Linear Inequalities

Linear Inequalities:

Solving linear inequalities is as easy as solving linear equations. You only need to know the following principles.

For a given inequality,

1. Adding the same number to each side of the inequality does not change the symbol $<$.
2. Subtracting the same number from each side of the inequality does not change the symbol $<$.
3. Multiplying or dividing the inequality by the same positive number does not change the symbol $<$.
4. Multiplying or dividing the inequality by the same negative number reverses the symbol $<$ from, say, $<$ to $>$.

The principle 4 can be easily understood with a simple example, say everybody would agree that $1<2$. If one multiplies the inequality by $-1$, it is true that $-1>-2$, i.e. the symbol $<$ reverses.

Example. Solve the inequality
$$3x-5<6-2x.$$

Solution. Adding 5 to each side of the inequality results
$$3x<11-2x.\ \ \ \ \ \mbox{(1)}$$
Adding $2x$ to each side of (1) results
$$5x<11.\ \ \ \ \ \mbox{(2)}$$
Dividing the inequality (2) by the same positive number 5 results
$$x<\frac{11}{5}.$$
There are various ways to write the solution. Normally $x<\frac{11}{5}$ would suffice but it can be written more formally as the solution set
$$\left\{x|\ x<\frac{11}{5}\right\}.$$
In interval notation, the solution can be written as
$$\left(-\infty,\frac{11}{5}\right).$$
Graphically it can be represented as

Example. Solve the inequality
$$13-7x\geq 10x-4.$$

Solution. Subtracting 13 from each side of the inequality results
$$-7x\geq 10x-17.\ \ \ \ \ \mbox{(3)}$$
Subtracting $10x$ from each side of (3) results
$$-17x\geq -17.\ \ \ \ \ \mbox{(4)}$$
Dividing each side of (4) by the same negative number $-17$ results
$$x\leq 1.$$

Note that the symbol $\geq$ has been reversed to $\leq$. The solution set can be written as $\{x|\ x\leq 1\}$ or $(-\infty,1]$ in interval notation. Graphically it can be represented as

Compound Inequalities:

A compound inequality is two inequalities joined by a conjunction AND or OR.

Example. Solve $-3<2x+5\leq 7$.

Solution. Notice that the compund inequality is formed by
$$-3<2x+5\ \mbox{and}\ 2x+5\leq 7.$$
Subtract 5 from each side of the given inequality.
$$-8<2x\leq 2.\ \ \ \ \ \mbox{(5)}$$
Devide (5) by 2.
$$-4<x\leq 1.$$
Hence the solution set is $\{x|\ -4<x\leq 1\}$ or $(-4,1]$ in interval notation. The graph of the solution set is given by

Example. Solve $2x-5\leq -7$ or $2x-5>1$.

Solution. Add 5 to eqch side of the two inequalities:
$$2x\leq -2\ \mbox{or}\ 2x>6.\ \ \ \ \ \mbox{(6)}$$
Divide each side of the two inequalities (6) by 2:
$$x\leq -1\ \mbox{or}\ x>3.$$
The solution set is $\{x|\ x\leq -1\ \mbox{or}\ x>3\}$ or $(-\infty,-1]\cup(3,\infty)$ in interval notation. The graph of the solution set is given by

# MAT 101 Online Lecture Notes: 2.5 More Equation Solving

Rational Equations:

The Recipe of Solving Rational Equations

1. First find the LCD (Least Common Denominator), i.e. the Least Common Multiple of all denominators.
2. Multiply your rational equation by the LCD.
3. Solve the resulting equation (usually a linear equation or a quadratic equation).
4. Plug solutions you obtained from STEP 3 into the original rational equation for $x$ to see if they all satisfy.

Example. Solve
$$\frac{1}{3x+6}-\frac{1}{x^2-4}=\frac{3}{x-2}.$$

Solution. STEP 1. $3x+6=3(x+2)$ and $x^2-4=(x+2)(x-2)$. Thus the LCD is $3(x+2)(x-2)$.

STEP 2. Multiply the equation the by LCD.
$$\frac{1}{3x+6}\cdot 3(x+2)(x-2)-\frac{1}{x^2-4}\cdot 3(x+2)(x-2)=\frac{3}{x-2}\cdot 3(x+2)(x-2)$$
which results the linear equation.
$$(x-2)-3=3\cdot 3(x+2).$$
The solution of this linear equation is $x=-\frac{23}{8}$.

The Recipe of Solving Radical Equations

1. First isolate one radical term in one side.
2. Square both sides of the equation.
3. If all radical are gone, solve the resulting equation (usually linear or quadratic). If not (in case the radical equation had two radical terms), repeat the steps 1 and 2.
4. Plug solutions you obtained from STEP 3 into the original radical equation for $x$ to see if they all satisfy.

Example. Solve $5+\sqrt{x+7}=x$.

Solution. STEP 1. Isolate the radical term $\sqrt{x+7}$ in the LHS.
$$\sqrt{x+7}=x-5.$$

STEP 2. Square both sides of the resulting equation.
$$x+7=(x-5)^2.$$
This is simplified to the quadratic equation
$$x^2-11x+18=0$$ which is factored to
$$(x-9)(x-2)=0.$$
Hence we obtain the two solutions $x=2,9$. However, not all these solutions may satisfy the original radical equation.

STEP 3. If $x=9$, then $$\mbox{LHS}=5+\sqrt{9+7}=9=\mbox{RHS}.$$ However, if $x=2$ then
$$\mbox{LHS}=5+\sqrt{x+7}=8\ne 2=\mbox{RHS}.$$ Therefore $x=9$ is the only solution to the radical equation.

Example. Solve $\sqrt{x-3}+\sqrt{x+5}=4$.

Solution. STEP 1. Isolate on radical term in one side, say isolate $\sqrt{x-3}$ in the LHS.
$$\sqrt{x-3}=4-\sqrt{x+5}.$$

STEP 2. Square both side of the resulting equation.
\begin{align*}
x-3&=(4-\sqrt{x+5})^2\\
&=16-8\sqrt{x+5}+(x+5)\\
&=x+21-8\sqrt{x+5}.
\end{align*}

Repeat STEP 1. Isolate the radical term $\sqrt{x+5}$ in the RHS.
$$3=\sqrt{x+5}.$$

Repeat STEP 2. Square both sides of the resulting equation.
$$9=x+5.$$
Hence we obtain the solution $x=4$.

One can readily check that $x=4$ satisfies the original radical equation.

# MAT 101 Online Lecture Notes: 2.4 Analyzing Graphs of Quadratic Functions

There are two important topics in this section: graphing the quadratic function $f(x)=ax^2+bx+c$ and finding the (absolute) maximum or the minimum value of $f(x)=ax^2+bx+c$.

First the sign of the leading coefficient $a$ tells us some information about the graph. If $a>0$ then the tail of the graph goes up, i.e. the graph is a smiling face $\smile$. If $a<0$ then the tail of the graph goes down, i.e. the graph is a frowning face $\frown$.

Using the completing the square $f(x)=ax^2+bx+c$ can be written as
$$f(x)=a(x-h)^2+k,$$
where $h=-\frac{b}{2a}$ and $k=f(h)=f\left(-\frac{b}{2a}\right)$. The ordered pair $\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)$ is called the vertex of the parabola $f(x)$ and the vertical line $x=-\frac{b}{2a}$ is called the axis of symmetry (this is the vertical line that divides the graph of $f(x)$ into two halves). If $a>0$, then $f\left(-\frac{b}{2a}\right)$ is the absolute minimum value of $f(x)$. If $a<0$, then $f\left(-\frac{b}{2a}\right)$ is the absolute maximum value of $f(x)$.

How to sketch the graph of $f(x)=a(x-h)^2+k$?

Your textbook is telling you to sketch the graph of $f(x)=a(x-h)^2+k$ using transformations that you learned in section 1.7 (here and here). In principle, it is right to use transformations but in practice there is an easier way to do. All you need is the sign of $a$, the vertex $\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)$, and the $y$-intercept $c$. (Although not required, it would be better if you know $x$-intercepts as well.)

Example. Let $f(x)=x^2+7x-8$.

(a) Find the vertex.

Solution. $-\frac{b}{2a}=-\frac{7}{2}$ and
\begin{align*}
f\left(-\frac{b}{2a}\right)&=f\left(-\frac{7}{2}\right)\\
&=\left(-\frac{7}{2}\right)^2+7\left(-\frac{7}{2}\right)-8\\
&=-\frac{81}{4}.
\end{align*}

(b) Find the axis of symmetry.

Solution. The axis of symmetry is the vertical line $x=-\frac{b}{2a}=-\frac{7}{2}$.

(c) Determine whether there is a maximum or minimum value and find that value.

Solution. Since $a=1>0$, there is a minimum and the minimum value is the $y$-coordinate of the vertex $f\left(-\frac{7}{2}\right)=-\frac{81}{4}$.

(d) Graph the function.

Solution. Since $a=1>0$, the graph is a parabola that opens up (smiling face). Also note that the $y$-intercept of $f(x)$ is $-8$. In fact, we can extract more information since $f(x)$ can be easily factored as $(x+8)(x-1)$, so the $x$-intercepts are $x=-8,1$.

# MAT 101 Online Lecture Notes: 2.3 Quadratic Equations, Functions and Models

Main topic in this section is solving a quadratic equation $ax^2+bx+c=0$. There are three ways to solve a quadratic equation. The first one is

1. By Factoring: This is a typical method to solve a quadratic equation whenever the polynomial $ax^2+bx+c$ can be easily factored. Here is an example.

Example. Solve the quadratic equation $x^2-3x-4=0$ by factoring.

Solution. The polynomial $x^2-3x-4$ is factored as $(x-4)(x+1)$. So the equation is $(x-4)(x+1)=0$. This means that $x-4=0$ or $x+1=0$, i.e. we obtain two real solutions $x=-1$ or $x=4$.

Example. Solve the quadratic equation $x^2-3=0$.

Solution 1. Recall the factorization formula $(a^2-b^2)=(a+b)(a-b)$. Now
\begin{align*}
x^2-3&=x^2-(\sqrt{3})^2\\
&=(x+\sqrt{3})(x-\sqrt{3}).
\end{align*}
Thus our equation becomes $(x+\sqrt{3})(x-\sqrt{3})=0$ whose solutions are $x=\pm\sqrt{3}$.

Solution 2. The quadratic equation can be written as $x^2=3$. Solving this equation for $x$, we obtain $x=\pm\sqrt{3}$.

Next method is

2. By Completing the Square:

This is a method that can be used to solve any quadratic equation. First note that $$x^2+bx+\left(\frac{b}{2}\right)^2=\left(x+\frac{b}{2}\right)^2.\ \ \ \ \ \mbox{(1)}$$

Example. Solve the equation $x^2-6x-10=0$ by completing the square.

Solution. By adding 10 to each side of the equation, we obtain
$$x^2-6x=10.\ \ \ \ \ \mbox{(2)}$$ Note that half of the coefficient of $x$ is $\frac{-6}{2}=-3$. Add $(-3)^2$ to each side of (2):
$$x^2-6x+(-3)^2=10+(-3)^2.\ \ \ \ \ \mbox{(3)}$$ Now notice that the LHS of (3) is exactly the same form as the LHS of the forumla (1). Hence, the quation (3) becomes
$$(x-3)^2=19.$$ Solving this for $x-3$, we obtain $x-3=\pm\sqrt{19}$. That is, $x=3\pm\sqrt{19}$.

While completing the square can be a useful tool for some other things, I do not strongly recommend this method because there is more convenient method of solving quadratic equations.

If you apply the method by completing the square to solve the quadratic equation $ax^2+bx+c=0$, we obtain the quadratic formula
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\ \ \ \ \ \mbox{(4)}.$$

Example. Solve the quadratic equation $3x^2+2x-7=0$.

Solution. $a=3$, $b=2$, and $c=-7$. Thus
\begin{align*}
x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\
&=\frac{-2\pm\sqrt{2^2-4(3)(-7)}}{2(3)}\\
&=\frac{-1\pm\sqrt{22}}{3}.
\end{align*}

The expression inside radical $b^2-4ac$ is called the discriminant. Using the discriminant, we can tell the following without solving the equation itself.

Theorem. For $ax^2+bx+c=0$ with $a\ne 0$,

• If $b^2-4ac>0$, then the equation has two distinct real solutions.
• If $b^2-4ac=0$, then the equation has only one real solution (which is $x=-\frac{b}{2a}$).
• If $b^2-4ac<0$, then the equation has two complex solutions that are conjugate of each other.

# MAT 101 Online Lecture Notes: 1.7 Symmetry and Transformations II

In this notes, we study the remaining transformations of a function, namely horizontal shifting, vertical shifting, and dilation (stretching and shrinking). We also discuss how those transformations can be used to sketch the graph of a complicated function from its most basic form.

Horizontal Shifting (translation to the right or left):

Example. Let us plot $y=x^2$, $y=(x-3)^2$, and $y=(x+3)^2$ all together. As you can see clearly, the graph of $y=(x-3)^2$ is 3 units translation to the right of the graph of $y=x^2$, while the graph of $y=(x+3)^2$ is 3 units translations to the left of the graph of $y=x^2$.

Let $a>0$. Then:

The graph of $y=f(x-a)$ is $a$ units translation to the right of the graph of $y=f(x)$;

The graph of $y=f(x+a)$ is $a$ units translation to the left of the graph of $y=f(x)$.

Vertical Shifting (translation upward or downward):

Example. Let us plot $y=x^2$, $y=x^2+3$, and $y=x^2-3$ all together. As you can see clearly, the graph of $y=x^2+3$ is 3 units translation upward of the graph of $y=x^2$, while the graph of $y=x^2-3$ is 3 units translations downward of the graph of $y=x^2$.

Let $a>0$. Then:

The graph of $y=f(x)+a$ is $a$ units translation to upward of the graph of $y=f(x)$;

The graph of $y=f(x)-a$ is $a$ units translation to downward of the graph of $y=f(x)$.

Stretching and Shrinking:

Example. Let us compare the graphs of $y=x^2$, $y=2x^2$ and $y=\frac{1}{2}x^2$. The graph of $y=2x^2$ looks thinner than the graph of $y=x^2$, while the graph of $y=\frac{1}{2}x^2$ looks wider than the graph of $y=x^2$. The graph of $y=2x^2$ can be considered as a (vetical) stretching of the graph of $y=x^2$. Similarly, the graph of $=\frac{1}{2}x^2$ can be considered as a (vetical) shrinking of the graph of $y=x^2$.

If $a>0$, the graph of $y=af(x)$ is a (vertical) stretching of the graph of $y=f(x)$;

If $0<a<1$, the graph of $y=af(x)$ is a (vertical) shrinking of the graph of $y=f(x)$.

Graphing functions using transformations: Sometimes we can sketch graph of a function by analyzing transformations involved such as reflection, stretching/shrinking, horizontal/vertical shifting.

Example. Sketch the graph of $f(x)=\frac{1}{3}x^3+2$ using transformations.

Solution. The most basic form of the function is $y=x^3$. So we obtain the graph of $f(x)$ from $y=x^3$

1. first by shrinking $y=\frac{1}{3}x^3$
2. and then translate the resultign function 2 units upward $y=\frac{1}{3}x^3+2$.

Example. Sketch the graph of $g(x)=-(x-3)^2+5$ using transformations.

Solution. The most basic form of the function is $y=x^2$. We obtain the graph of $g(x)$ from $y=x^2$

1. first by reflection $y=-x^2$
2. next translate the resulting function 3 units to the right
3. and finally translate the resulting function 5 units upward.

# MAT 101 Online Lecture Notes: 1.7 Symmetry and Transformations I

Transformations of the graph of a function $y=f(x)$:

The transformations we consider include reflection, horizontal shifting, vertical shifting, stretching and shrinking. Here we discuss reflection and related symmetries. In the following notes, we will discuss the rest of transformations.

Reflection about the $x$-axis:

Example. Consider $y=\sqrt{x}$. If you plug in $-y$ for $y$, we obtain $-y=\sqrt{x}$ or $y=-\sqrt{x}$. Let us plot them all togather.

If you fold the two plots along the $x$-axis, the graphs of $y=\sqrt{x}$ and $y=-\sqrt{x}$ would overlap. So we can say that $y=-\sqrt{x}$ is the reflection of $y=\sqrt{x}$ with respect to $x$-axis, and vice versa.

In general, the graph of $y=-f(x)$ is the reflection of the graph of $y=f(x)$ with respect to the $x$-axis.

Reflection about the $y$-axis:

Example: Now this time we consider the function $y=2x+1$. If you plug in $-x$ for $x$, we obtain $y=-2x+1$. Let us plot them both again.

You can clearly see that if you fold the two plots along teh $y$-axis, they would overlap. So we can say that the graph of $y=-2x+1$ is the reflection of the graph of $y=2x+1$.

In general, the graph of $y=f(-x)$ is the reflection of the graph of $y=f(x)$ with respect to the $y$-axis.

Symmetry about the $y$-axis:

Example. Consider the function $f(x)=x^2-1$. If you plug in $-x$ for $x$, $f(-x)=(-x)^2-1=x^2-1=f(x)$. So the function is unchanged by the reflection with respect to the $y$-axis.

In general, if the function $y=f(x)$ satisfies the property $$f(-x)=f(x),$$ we say that the graph of $f(x)$ is symmetric about the $y$-axis.

Symmetry about the $x$-axis:

Example: Consider $x=y^2$. This is a function of the form $x=f(y)$. If you plug in $-y$ for $y$, $(-y)^2=y^2=x$. So the function still does not change.

In general, if the function $x=f(y)$ satisfies the property $$f(-y)=f(y),$$ we say that the graph of $x=f(y)$ is symmetric about the $x$-axis.

Example: Consider the function $y=x^3$. If you plug in $-x$ for $x$ and $-y$ for $y$, the equation does not change.

The effect of $x\mapsto -x$ and $y\mapsto -y$ in the equation $y=f(x)$ amounts to rotating the graph of $y=f(x)$ about the origin by $180^\circ$. If the graph of $y=f(x)$ does not change after the rotation of the graph about the origin by $180^\circ$ (or equivalently $-y=f(-x)$ is the same as the equation $y=f(x)$), then we say that the graph is symmetric about the origin.

Even Functions and Odd Functions:

Definition. A function $y=f(x)$ is a called an even function if it satisfies $f(-x)=f(x)$. The reason such function is called even is that one of the simplest examples of even functions is of the form $x^{\mbox{even integer}}$, for instance, $x^2$, $x^4$, $x^6$, etc.

Definition. A function $y=f(x)$ is a called an odd function if it satisfies $f(-x)=-f(x)$. The reason such function is called odd is that one of the simplest examples of odd functions is of the form $x^{\mbox{odd integer}}$, for instance, $x$, $x^3$, $x^5$, etc.

From the equations we can easily see that the graph of an even functions is symmetric about the $y$-axis, while the graph of an odd function is symmetric about the origin.

Example. $f(x)=5x^7-6x^3-2x$ is an odd function because \begin{align*}f(-x)&=5(-x)^7-6(-x)^3-2(-x)\\&=-5x^7+6x^3+2x\\&=-f(x).\end{align*}

Remark. Note that a constant multiple of an odd function is an odd function and the sum of an odd functions is an odd function. Since each term $x^7$, $x^3$, $x$ are odd functions, we can also see that $f(x)$ is an odd function.

Example. $g(x)=5x^6-3x^2-7$ is an even function because \begin{align*}
g(-x)&=5(-x)^6-3(-x)^2-7\\&=5x^6-3x^2-7\\&=g(x).
\end{align*}

Remark. Note that a constant function (number) is an even function, a constant multiple of an even function is an even function and the sum of an even functions is an even function. Since each term $x^6$, $x^2$, $-7$ are even functions, we can also see that $g(x)$ is an even function.