Period 4: Solving Inequalities Algebraically

Inequality deals with a range of values, rather than an exact number, so it is necessary to replace the equal sign in an equation with its supposed inequality (either >, <, >, or <).

For example, Solve for the following inequality:

(x + 10) (2x – 9) > 0

1. Break up both of the brackets into two separate “sections”, x +12>0 and 2x - 9 >0.
2. So far, we know that each individual section must exceed the value of 0, as this is the inequality. From this, we can isolate the x value to determine the roots.
   
   x+ 12 > 0
   x > -10
   
   2x-9 >0
   = 2x> 9
   x > 9/2
   
   Note that for section 2, the value of x can also be 0.
   
   So now we know that the roots of the equation when (x + 10) (2x – 9) = 0 are x = -10, x = 9/2, and x = 0
   
   

3. We can then break up number line into three intervals in relation to the roots that we have found.
   
   Because the lowest root is -10, we recognize in order for the graph to continue negatively, x must therefore be less than -10, so x < -10.
   
   Likewise, the highest root value is 9/2 so in order for the graph to continue positively, x must be bigger than 9/2, so 9/2 > x. We must also notice that the middle interval must (graphically) be larger than -10, yet smaller than 9/2, so, respectively, the inequality is represented by -10 <>
   
   
   
4. You can test out each interval by plugging it into the equation and determining which one is the solution.
   
   For x<-10, test x = -11
   (-11 + 10) (2 (-11) -9) = 31
   because the value “31” is greater than “0”, x< -10 is a solution. For -10 <9/2,>x = 1
   (1+10) (2(1)-9) = -77
   because the value “-77” is less than “0”, -10 <> solution. For x > 9/2, test x = 5
   (5+10) (2(5)-9) = 135
   because the value “135” is greater than “0”, x>9/2 is a solution.
   
   
5. Summarize the information found onto a table:
   

   Interval Factor	x < -10	-10	x > 9/2
   (x +10)	( - )	( + )	( + )
   (2x – 9)	( - )	( - )	( + )
   f(x)	( + )	( - )	( + )

   
   
6. The solution, shown on the number line, is x < -10 and x > 9/2

   

   Note: Page 8 of our textbooks summarizes all possible intervals
   
   

   Therefore, the solution is XE(-∞, -10] U[9/2, ∞)

2.4: Families of Polynomial Functions

Hello fellow students, I’m sorry for the delayed post for Ms. Burchat’s third period class.

My scribe post was Friday’s lesson, lesson 2.4: Families of Polynomial Functions. A good way to think of families of polynomial functions is like flowers. How flowers you say? Well think of it this way, if you choose one type of flower; for instance a rose, all roses are typically the same making them a family. However what differentiates between them are their colours and sizes, this would specify or identify a specific rose from the family of roses.

So how do roses relate to functions? Now that you understand what is meant by a family you can understand families of functions. When we say families of polynomial functions we mean to say in a linear, quadratic, cubic, quartic or quintic function these would be know as your families and to differentiate from one quadratic to another it would depend on the numerical values of each quadratic function which would make that individual function a specific one from the family of quadratic functions.

In Friday’s class we learned to do so algebraically, here is an example.

If we say we have the zeros of a family for cubic functions are 4, -1, and 3

Then we say:
a) Determine an equation for this family.
b) Determine an equation for the member of the family whose graph has a y-intercept of 17.

Solution:
a) Since the zeros are: 4, -1, and 3. (x-4)(x+1)(x-3) are also factors of the family of cubics. Therefore the equation for this question would be simply just to equate the factors to 0 and solve for x, as follows:

x-4= 0, then x= 4
x+1=0, then x= -1
x-3=0, then x=3
So, now we write the equation:
y= (x-4)(x+1)(x-3)

b) Now, we sub in the y-intercept in order to find the value of K. Using the equation we created above: y=k (x-4)(x+1)(x-3), we will put instead of y, 17 and 0’s for the x- intercepts.

Make note that we are using 0’s for the x-intercept and not an actually value because in the question it did not give us a pair of points that passed through the function like follows: (5,17). In this case we would sub in X with 5 and y with 17. But since we are only given the y- intercept we assume x is 0.

So with the function: y= k(x-4)(x+1)(x-3)

17= k(0-4)(0+1)(0-3)
17= -4k+1-3
17=2k
17/ 2=k
8.5=k
And therefore, y= 8.5(0-4)(0+1)(0-3)
This would be a SPECIFIC cubic function from the many members of the family of cubics.