# 1.2 Basic classes of functions  (Page 4/28)

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Now consider a cubic function $f\left(x\right)=a{x}^{3}+b{x}^{2}+cx+d.$ If $a>0,$ then $f\left(x\right)\to \infty$ as $x\to \infty$ and $f\left(x\right)\to \text{−∞}$ as $x\to \text{−∞}.$ If $a<0,$ then $f\left(x\right)\to \text{−∞}$ as $x\to \infty$ and $f\left(x\right)\to \infty$ as $x\to \text{−∞}.$ As we can see from both of these graphs, the leading term of the polynomial determines the end behavior. (See [link] (b).)

## Zeros of polynomial functions

Another characteristic of the graph of a polynomial function is where it intersects the $x$ -axis. To determine where a function $f$ intersects the $x$ -axis, we need to solve the equation $f\left(x\right)=0$ for .n the case of the linear function $f\left(x\right)=mx+b,$ the $x$ -intercept is given by solving the equation $mx+b=0.$ In this case, we see that the $x$ -intercept is given by $\left(\text{−}\mathit{\text{b}}\text{/}m,0\right).$ In the case of a quadratic function, finding the $x$ -intercept(s) requires finding the zeros of a quadratic equation: $a{x}^{2}+bx+c=0.$ In some cases, it is easy to factor the polynomial $a{x}^{2}+bx+c$ to find the zeros. If not, we make use of the quadratic formula.

## Rule: the quadratic formula

Consider the quadratic equation

$a{x}^{2}+bx+c=0,$

where $a\ne 0.$ The solutions of this equation are given by the quadratic formula

$x=\frac{\text{−}b±\sqrt{{b}^{2}-4ac}}{2a}.$

If the discriminant ${b}^{2}-4ac>0,$ this formula tells us there are two real numbers that satisfy the quadratic equation. If ${b}^{2}-4ac=0,$ this formula tells us there is only one solution, and it is a real number. If ${b}^{2}-4ac<0,$ no real numbers satisfy the quadratic equation.

In the case of higher-degree polynomials, it may be more complicated to determine where the graph intersects the $x$ -axis. In some instances, it is possible to find the $x$ -intercepts by factoring the polynomial to find its zeros. In other cases, it is impossible to calculate the exact values of the $x$ -intercepts. However, as we see later in the text, in cases such as this, we can use analytical tools to approximate (to a very high degree) where the $x$ -intercepts are located. Here we focus on the graphs of polynomials for which we can calculate their zeros explicitly.

## Graphing polynomial functions

For the following functions a. and b., i. describe the behavior of $f\left(x\right)$ as $x\to \text{±}\infty ,$ ii. find all zeros of $f,$ and iii. sketch a graph of $f.$

1. $f\left(x\right)=-2{x}^{2}+4x-1$
2. $f\left(x\right)={x}^{3}-3{x}^{2}-4x$
1. The function $f\left(x\right)=-2{x}^{2}+4x-1$ is a quadratic function.
1. Because $a=-2<0,\text{as}\phantom{\rule{0.2em}{0ex}}x\to \text{±}\infty ,f\left(x\right)\to \text{−∞.}$
2. To find the zeros of $f,$ use the quadratic formula. The zeros are
$x=\frac{-4±\sqrt{{4}^{2}-4\left(-2\right)\left(-1\right)}}{2\left(-2\right)}=\frac{-4±\sqrt{8}}{-4}=\frac{-4±2\sqrt{2}}{-4}=\frac{2±\sqrt{2}}{2}.$
3. To sketch the graph of $f,$ use the information from your previous answers and combine it with the fact that the graph is a parabola opening downward.
2. The function $f\left(x\right)={x}^{3}-3{x}^{2}-4x$ is a cubic function.
1. Because $a=1>0,\text{as}\phantom{\rule{0.2em}{0ex}}x\to \infty ,f\left(x\right)\to \infty .$ As $x\to \text{−∞},f\left(x\right)\to \text{−∞}.$
2. To find the zeros of $f,$ we need to factor the polynomial. First, when we factor $x$ out of all the terms, we find
$f\left(x\right)=x\left({x}^{2}-3x-4\right).$

Then, when we factor the quadratic function ${x}^{2}-3x-4,$ we find
$f\left(x\right)=x\left(x-4\right)\left(x+1\right).$

Therefore, the zeros of $f$ are $x=0,4,-1.$
3. Combining the results from parts i. and ii., draw a rough sketch of $f.$

#### Questions & Answers

questions solve y=sin x
Solve it for what?
Tim
you have to apply the function arcsin in both sides and you get arcsin y = acrsin (sin x) the the function arcsin and function sin cancel each other so the ecuation becomes arcsin y = x you can also write x= arcsin y
Ioana
what is the question ? what is the answer?
Suman
there is an equation that should be solve for x
Ioana
ok solve it
Suman
are you saying y is of sin(x) y=sin(x)/sin of both sides to solve for x... therefore y/sin =x
Tyron
or solve for sin(x) via the unit circle
Tyron
what is unit circle
Suman
a circle whose radius is 1.
Darnell
the unit circle is covered in pre cal...and or trigonometry. it is the multipcation table of upper level mathematics.
Tyron
what is function?
A set of points in which every x value (domain) corresponds to exactly one y value (range)
Tim
what is lim (x,y)~(0,0) (x/y)
limited of x,y at 0,0 is nt defined
Alswell
But using L'Hopitals rule is x=1 is defined
Alswell
Could U explain better boss?
emmanuel
value of (x/y) as (x,y) tends to (0,0) also whats the value of (x+y)/(x^2+y^2) as (x,y) tends to (0,0)
NIKI
can we apply l hospitals rule for function of two variables
NIKI
why n does not equal -1
ask a complete question if you want a complete answer.
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.
proof the formula integration of udv=uv-integration of vdu.?
Find derivative (2x^3+6xy-4y^2)^2
no x=2 is not a function, as there is nothing that's changing.
are you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.
The
i mean can we replace the roles of x and y and call x=2 as function
The
if x =y and x = 800 what is y
y=800
800
Bg
how do u factor the numerator?
Nonsense, you factor numbers
Antonio
You can factorize the numerator of an expression. What's the problem there? here's an example. f(x)=((x^2)-(y^2))/2 Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y). so the overall function becomes : ((x+y)(x-y))/2
The
The problem is the question, is not a problem where it is, but what it is
Antonio
I think you should first know the basics man: PS
Vishal
Yes, what factorization is
Antonio
Antonio bro is x=2 a function?
The
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
Antonio
you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though
zach
Why y, if domain its usually defined as x, bro, so you creates confusion
Antonio
Its f(x) =y=2 for every x
Antonio
Yes but he said could you put x = 2 as a function you put y = 2 as a function
zach
F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)
Antonio
x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent
zach
The said x=2 and that 2 is y
Antonio
that 2 is not y, y is a variable 2 is a constant
zach
So 2 is defined as f(x) =2
Antonio
No y its constant =2
Antonio
what variable does that function define
zach
the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2
zach
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Antonio
Sorry x=2
Antonio
And you are right, but os not a function of x, its a function of y
Antonio
As function of x is meaningless, is not a finction
Antonio
yeah you mean what I said in my first post, smh
zach
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
Antonio
OK you can call this "function" on a set {2}, but its a single value function, a constant
Antonio
well as long as you got there eventually
zach
2x^3+6xy-4y^2)^2 solve this
femi
follow algebraic method. look under factoring numerator from Khan academy
moe
volume between cone z=√(x^2+y^2) and plane z=2
Fatima
It's an integral easy
Antonio
V=1/3 h π (R^2+r2+ r*R(
Antonio
How do we find the horizontal asymptote of a function using limits?
Easy lim f(x) x-->~ =c
Antonio
solutions for combining functions
what is a function? f(x)
one that is one to one, one that passes the vertical line test
Andrew
It's a law f() that to every point (x) on the Domain gives a single point in the codomain f(x)=y
Antonio
is x=2 a function?
The
restate the problem. and I will look. ty
is x=2 a function?
The