FunctionsMAT 102

Example3.B.5Polynomial Addition

Add the two cubic polynomials.

Example3.B.6Polynomial Subtraction

Subtract the two quadratic polynomials.

Example3.B.7Polynomial Addition and Subtraction

Simplify.

Example3.B.8Polynomial Multiplication

Multiply $(4x + 3) (3x - 2).$

Example3.B.9Polynomial Multiplication

Multiply $(x - 2) (4x^2 - x +5).$

Example3.B.10Polynomial Multiplication (Reverse Order)

Multiply $(4x^2 - x +5)(x - 2) .$

Example3.B.11Graph Quadratic Polynomial

Consider the graph of $f(x) = x^2+1\text{.}$ We'll use the low tech but effective method of plotting points:

$x$	$y$
$-2$	$5$
$-1$	$2$
$0$	$1$
$1$	$2$
$2$	$5$

The $y$-intercept is at $y=1$ and there is no $x$-intercept since the equation $0 = x^2+1$ has no solution that is a real number. Notice that if we plot points for values of $x$ further to the left of $-2\text{,}$ the function will produce large positive $y$ values since any negative number squared is positive. If we try $x$ values to the right of $2\text{,}$ we clearly get large positive values for $y\text{.}$ We are confident this graph extends upward in both directions.$\checkmark$

Example3.B.12Graph Quadratic Polynomial

Consider the graph of $g(x) = -x^2-x+2\text{.}$

The $y$-intercept is easy to find:

The $x$-intercept takes a little more work: How to find when $0 =-x^2-x+2\text{?}$ Here's when all that time you spent learning how to factor comes in handy. We won't do a lot of factoring in this course and the "guess and check" method will be sufficient.

$-2$ and $1$ don't work, so try $2$ and $-1\text{:}$

Now, back to the equation: $0 =x^2+x-2= (x+2)(x-1)\text{.}$ The only way to get zero when two numbers are multiplied is for one of the numbers to be zero:

To see what the “end behavior” is, note that the leading term produces values of much greater magnitude than the other terms in the polynomial. For example, when $x = 10\text{,}$ the leading term makes: $-(10)^2=-100\text{.}$ The rest of the polynomial only contributes, $-(10)+2 = -8\text{.}$ If we go just a little further out, say $x = 15\text{,}$ compare the leading term's value to the rest of the polynomial: $-(15)^2=-225$ versus $-(15)+2 = -13\text{.}$ What about the graph for negative values of $x\text{?}$ Check out: $-(-15)^2=-225$ versus $-(-15)+2 = 17\text{.}$ We can see that the leading term will produce negative $y$-values for $x$ values to the far left or right. Our analysis shows that both ends of $g(x)$ are downward having negative $y$ values.

Plotting a few more points gives the graph:

Example3.B.13Graph Cubic Polynomial

Consider the graph $h(x) = \frac{2}{7}x^3-\frac{1}{7}x^2-3x.$

It's easy to see that the $y$-intercept is at $y=0\text{.}$ We need to factor the polynomial to find the $x$-intercepts. Here's a "tip of the day": factoring is easier if you factor out the denominators and anything else the terms have in common. Notice below that when we factor out the $\frac{1}{7}\text{,}$ we have to adjust the $3$ so that if the $\frac{1}{7}$ were multiplied back in, we'd still have $3\text{.}$

So,

For the end behavior: The leading term is $\frac{2}{7}x^3\text{,}$ the power $3$ will make negative $y$ values for negative $x$ values and positive $y$ values for positive $x$ values. The left end will be “down” and the right end will go “up”.

Plotting a few more points gives the graph:

Example3.B.14Graph Fourth-Degree Polynomial

Consider the graph $k(x) = x^4-2x^3-2x^2.$

The $y$-intercept is at $y=0\text{,}$ but we have a problem when trying to factor to find the $x$-intercepts:

The only two combinations that multiply to get $-2$ for a "nice" factorization are $1$ and $-2\text{,}$ or $-1$ and $2\text{.}$ Checking we see that

But neither gives $x^2 - 2x - 2\text{.}$ Luckily, for quadratic polynomials, there is a formula which will give the solutions to equations like, $x^2 - 2x - 2 = 0\text{,}$ the Quadratic Formula. We pause to recall what it says:

Quadratic Formula For $ax^2 + bx + c = 0\text{,}$ the solutions are found by

Now we may breathe easier, knowing we can still find the $x$-intercepts even if we can't factor the polynomial. (By the way, the Quadratic formula works even if the polynomial factors nicely.)

Back to the graph: So far, we had $x^4-2x^3-2x^2 = x^2(x^2 - 2x - 2)$ which means

and clearly, $x=0$ is one of the $x$-intercepts. Now, consider the “not so easy” to factor part: $x^2 - 2x - 2 = 0\text{.}$ Here, $a=1\text{,}$ $b=-2$ and $c=-2\text{.}$ With help from the Quadratic Formula:

and using a calculator: $x = \frac{2 + \sqrt{12}}{2} \approx 2.732$ or $x = \frac{2 - \sqrt{12}}{2} \approx -0.732\text{.}$ At last! We have the $x$-intercepts and we only need to determine the end behavior. The leading term is $x^4$ and any number raised to the $4$-th power produces a positive number. Hence, both ends are upward.

Plot a few more points and we are done:

Example3.B.15Quadratic Formula

We should probably work a few problems with the Quadratic Formula, just for practice:

1. Solve: $x^2 + 3x = 1$

   First, we must rewrite this as an equivalent equation equalling $0\text{.}$ We subtract $1$ from both sides of the equation.

   $$\begin{align*} x^2 + 3x \amp = 1\\ x^2 + 3x - 1 \amp = 0 \end{align*}$$

   With just a few tries, we can see that we are not going to make a $3$ from integers that multiply to get $-1\text{.}$ This indicates we must use the Quadratic Formula: $ax^2 + bx + c = 0\text{.}$ For this particular equation we have $a=1\text{,}$ $b=3$ and $c=-1\text{:}$

   $$\begin{align*} x\amp = \frac{-3 \pm \sqrt{3^2 - 4(1)(-1)}}{2(1)}\\ \amp = \frac{-3 \pm \sqrt{9 +4}}{2}\\ \amp = \frac{-3 \pm \sqrt{13}}{2} \end{align*}$$

   Remember that there are two values here. We wrote $\pm$ as a short-cut. The answers are: $x = \frac{-3 + \sqrt{13}}{2}$ or $x = \frac{-3 - \sqrt{13}}{2}.$

2. Solve: $x^2 + x + 5 = 0$

   Note that $a = 1\text{,}$ $b = 1$ and $c = 5:$

   $$\begin{align*} x\amp = \frac{-1 \pm \sqrt{1^2 - 4(1)(5)}}{2(1)}\\ \amp = \frac{-1 \pm \sqrt{1 - 20}}{2}\\ \amp = \frac{-1 \pm \sqrt{-19}}{2} \end{align*}$$

   But, there is no real number that squares itself to get $-19.$ Since the Quadratic Formula always finds the solution to a quadratic equation, there must not be any solutions for this equation. Our answer is: $\text{No solution}.$

3. Solve: $8x^3 - 24x^2 + 18x = 0$

   This equation is not a quadratic, but if we factor it we get: $2x(4x^2 - 12x + 9) = 0\text{.}$ One solution is $x = 0$ since we could have $2x=0\text{.}$ Using the quadratic formula for $4x^2 - 12x + 9=0$ with $a=4\text{,}$ $b=-12$ and $c=9\text{:}$

   $$\begin{align*} x\amp = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(4)(9)}}{2(4)}\\ \amp = \frac{12 \pm \sqrt{144 - 144}}{8}\\ \amp = \frac{12 \pm \sqrt{0}}{8}\\ \amp = \frac{12}{8}\\ \amp = \frac{3}{2} \end{align*}$$

   Hence, the solutions are $x = 0$ or $x = \frac{3}{2}\text{.}$ (Seems like we should have more solutions for all that work!)