Question

17. An iron rod is bent to the following shape for construction purpose
1) The above shape is a representation of
a) linear polynomial
b) quadratic polynomial
c) cubic polynomial
d) none of these
2) Write the number zeroes of graph of polynomial in the above graph
a)
b)3
c) 2 d) 5
3) Write the polynomial for the above representation
a) x +2x +1 b) x²-2x +1 c) x²2x-1
d) x²-1
4) Find the value of polynomial at x=2
a) 0
b)3
c) 2.
d) 5
5) Name the shape of the curve
a) parabola
b) hyperbola
c) oval
d) Elliptical​

Answer 1

The end behavior of a function fff describes the behavior of its graph at the "ends" of the xxx-axis. Algebraically, end behavior is determined by the following two questions:

As x\rightarrow +\inftyx→+∞x, right arrow, plus, infinity, what does f(x)f(x)f, left parenthesis, x, right parenthesis approach?

As x\rightarrow -\inftyx→−∞x, right arrow, minus, infinity, what does f(x)f(x)f, left parenthesis, x, right parenthesis approach?

If this is new to you, we recommend that you check out our end behavior of polynomials article.

The zeros of a function fff correspond to the xxx-intercepts of its graph. If fff has a zero of odd multiplicity, its graph will cross the xxx-axis at that xxx value. If fff has a zero of even multiplicity, its graph will touch the xxx-axis at that point.

If this is new to you, we recommend that you check out our zeros of polynomials article.

What you will learn in this lesson

In this lesson, we will use the above features in order to analyze and sketch graphs of polynomials. We will then use the sketch to find the polynomial's positive and negative intervals.

Analyzing polynomial functions

We will now analyze several features of the graph of the polynomial f(x)=(3x-2)(x+2)^2f(x)=(3x−2)(x+2)  

2

f, left parenthesis, x, right parenthesis, equals, left parenthesis, 3, x, minus, 2, right parenthesis, left parenthesis, x, plus, 2, right parenthesis, squared.

Finding the yyy-intercept

To find the yyy-intercept of the graph of fff, we can find f(0)f(0)f, left parenthesis, 0, right parenthesis.

\begin{aligned} f(x)&=(3x-2)(x+2)^2 \\\\ f(\tealD0)&= (3(\tealD 0)-2)(\tealD0+2)^2\\ \\ f(0)&= (-2)(4)\\\\ f(0)&=-8 \end{aligned}  

f(x)

f(0)

f(0)

f(0)

​  

 

=(3x−2)(x+2)  

2

 

=(3(0)−2)(0+2)  

2

 

=(−2)(4)

=−8

​  

 

The yyy-intercept of the graph of y=f(x)y=f(x)y, equals, f, left parenthesis, x, right parenthesis is (0,-8)(0,−8)left parenthesis, 0, comma, minus, 8, right parenthesis.

Finding the xxx-intercepts

To find the xxx-intercepts, we can solve the equation f(x)=0f(x)=0f, left parenthesis, x, right parenthesis, equals, 0.

\begin{aligned} f(x)&=(3x-2)(x+2)^2 \\\\ \tealD 0&= (3x-2)(x+2)^2\\ \\ \end{aligned}  

f(x)

0

​  

 

=(3x−2)(x+2)  

2

 

=(3x−2)(x+2)  

2

 

​  

 

\begin{aligned}&\swarrow&\searrow\\\\ 3x-2&=0&\text{or}\quad x+2&=0&\small{\gray{\text{Zero product property}}}\\\\ x&=\dfrac{2}{3}&\text{or}\qquad x&=-2\end{aligned}  

3x−2

x

​  

 

↙

=0

=  

3

2

​  

 

​  

 

↘

orx+2

orx

​  

 

=0

=−2

​  

 

Zero product property

​  

 

The xxx-intercepts of the graph of y=f(x)y=f(x)y, equals, f, left parenthesis, x, right parenthesis are \left(\dfrac23,0\right)(  

3

2

​  

,0)left parenthesis, start fraction, 2, divided by, 3, end fraction, comma, 0, right parenthesis and (-2,0)(−2,0)left parenthesis, minus, 2, comma, 0, right parenthesis.

Our work also shows that \dfrac 23  

3

2

​  

start fraction, 2, divided by, 3, end fraction is a zero of multiplicity 111 and -2−2minus, 2 is a zero of multiplicity 222. This means that the graph will cross the xxx-axis at \left (\dfrac 23, 0\right)(  

3

2

​  

,0)left parenthesis, start fraction, 2, divided by, 3, end fraction, comma, 0, right parenthesis and touch the xxx-axis at (-2,0)(−2,0)left parenthesis, minus, 2, comma, 0, right parenthesis.

Finding the end behavior

To find the end behavior of a function, we can examine the leading term when the function is written in standard form.

Let's write the equation in standard form.

\begin{aligned}f(x)&=(3x-2)(x+2)^2\\ \\ f(x)&=(3x-2)(x^2+4x+4)\\ \\ f(x)&=3x^3+12x^2+12x-2x^2-8x-8\\ \\ f(x)&=\goldD{3x^3}+10x^2+4x-8 \end{aligned}  f(x) f(x) f(x) f(x)

 

  =(3x−2)(x+2)  2

 =(3x−2)(x  2 +4x+4) =3x  3  +12x  2  +12x−2x  2  −8x−8 =3x  3  +10x  2 +4x−8 ​  

 

[Is it necessary to do this?]

The leading term of the polynomial is \goldD{3x^3}3x  

3

start color #e07d10, 3, x, cubed, end color #e07d10, and so the end behavior of function fff will be the same as the end behavior of 3x^33x  

3

3, x, cubed.

Since the degree is odd and the leading coefficient is positive, the end behavior will be: as x\rightarrow +\inftyx→+∞x, right arrow, plus, infinity, f(x)\rightarrow +\inftyf(x)→+∞f, left parenthesis, x, right parenthesis, right arrow, plus, infinity and as x\rightarrow -\inftyx→−∞x, right arrow, minus, infinity, f(x)\rightarrow -\inftyf(x)→−∞f, left parenthesis, x, right parenthesis, right arrow, minus, infinity.

Sketching a graph

We can use what we've found above to sketch a graph of y=f(x)y=f(x)y, equals, f, left parenthesis, x, right parenthesis.

Let's start with end behavior:

As x\rightarrow +\inftyx→+∞x, right arrow, plus, infinity, f(x)\rightarrow +\inftyf(x)→+∞f, left parenthesis, x, right parenthesis, right arrow, plus, infinity.

As x\rightarrow -\inftyx→−∞x, right arrow, minus, infinity, f(x)\rightarrow -\inftyf(x)→−∞f, left parenthesis, x, right parenthesis, right arrow, minus, infinity.