Exercise 6 Page 325

To sketch the graph of a function, we must find its zeros and determine how the function changes around these points.

Factoring the Polynomial

To find zeros of the given function, we need to solve the equation k(x)=0. To do this, we will need to define another variable. If we let z=x^2, we can rewrite the equation in terms of the z-variable. x^4-10x^2+9=0 ⇔ z^2-10z+9=0 Note that the above equation in terms of z is a quadratic equation. Thus, we can solve it using the Quadratic Formula. az^2+bz+c=0 ⇔ z=- b±sqrt(b^2-4ac)/2a Let's identify a, b, and c. z^2-10z+9=0 ⇔ 1z^2+( - 10)z+ 9=0We see that a= 1, b= - 10, and c= 9. Let's substitute these values into the formula, and solve for z.

z=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

z=- ( - 10)±sqrt(( - 10)^2-4( 1)( 9))/2( 1)

▼

Solve for z

NegNeg

- (- a)=a

z=10±sqrt((- 10)^2-4(1)(9))/2(1)

CalcPow

Calculate power

z=10±sqrt(100-4(1)(9))/2(1)

IdPropMult

Identity Property of Multiplication

z=10±sqrt(100-4(9))/2

MultNegPos

(- a)b = - ab

z=10±sqrt(100-36)/2

SubTerm

Subtract term

z=10±sqrt(64)/2

CalcRoot

Calculate root

z=10± 8/2

ReduceFrac

a/b=.a /2./.b /2.

z=5 ± 4

The solutions for this equation are z=5 ± 4. Let's separate them by using the positive and negative signs.

z=5 ± 4
z=5 + 4	z=5 - 4
z=9	z=1

We found that the solutions for z^2-10z+9=0 are z=9 and z=1. This means that x^2=9 and x^2=1. Let's solve these equations!

lcx^2=9 & (I) x^2=1 & (II)

(I), (II): sqrt(LHS)=sqrt(RHS)

lx=± 3 x=± 1

We found in total four solutions for x. x=2, x=- 3, x=1, x=- 1 Now we can rewrite given function in factored form. k(x)=(x-1)(x-(- 1))(x-3)(x-(- 3)) ⇕ k(x)=(x-1)(x+1)(x-3)(x+ 3)

Sketching the Graph

Having fully factored the function, we see that x=1, x=- 1, x=3, and x=- 3 are rational roots. Let's plot them in a coordinate plane.

If we know how the functions behaves around these zeros, we get an idea of what it looks like. To determine this, we can calculate the k-values in the following intervals. x&<- 3 - 3< x& < - 1 - 1< x& < 1 1< x& < 3 3< x& Let's choose some x-values in these intervals and find their corresponding k-values. The only x-value that will not be arbitrary is x=0, as we want to know where the graph intercepts the y-axis.

Interval	x	(x-1)(x+1)(x-3)(x+ 3)	k
x<- 3	- 4	( - 4-1)( - 4+1)( - 4-3)( - 4+ 3)	105
- 3< x < - 1	- 2	( - 2-1)( - 2+1)( - 2-3)( - 2+ 3)	- 15
- 1< x < 1	0	( 0-1)( 0+1)( 0-3)( 0+ 3)	9
1< x < 3	2	( 2-1)( 2+1)( 2-3)( 2+ 3)	- 15
3 < x	4	( 4-1)( 4+1)( 4-3)( 4+ 3)	105

With the exception of the y-intercept, the actual k-values for the given x-values we used are not important. Instead we are more interested if the function is above or below the x-axis in the given intervals. This will tell us how the function behaves. k(- 4)&= 105 &&⇒ Abovethex-axis k(- 2)& = - 15 &&⇒ Belowthex-axis k(0)&= 9 &&⇒ Abovethex-axis k(2)& = - 15 &&⇒ Belowthex-axis k(4)& = 105 &&⇒ Abovethex-axis Going from above the x-axis to below the x-axis means the function is decreasing, and vice-versa. With this information, we can sketch the graph. We will draw it so that it intercepts the y-axis at (0,9). From our randomly chosen substitutions, we also know that it should pass through (2,- 15). Adding these points to the diagram will help to make a better sketch.

Graphing the Function in a Calculator

Now let's graph the function in our graphing calculator and compare it to our sketch. Notice that we can use the same window-setting as in our sketch to make sure the proportions are the same.

The graphing calculator image is very similar to our graph.

Hints

Check the answer

Practice exercises

Asses your skill

Factoring the Polynomial

Sketching the Graph

Graphing the Function in a Calculator