Topic 1 and 2 Reteach

Topic 1 and 2 Reteach

Key Features of Functions

Functions in math are mathematical equations that have an independent variable (usually x) and a dependent variable (usually y). The x variable is the input or DOMAIN and the y variable is the output or RANGE.

Rate of Change

The ratio of change between the range (y) and the domain (x) is called the Rate of Change (ROC). All linear functions (x to the first power) have a ROC also called the slope of the line. To calculate the ROC simply divide the change in y by the change in x. Example: you are given two points (2, 3) and (5, 12) you find the change in y (9) and the change in x (3) and divide y's change by x's change (3)

 $m\ =\frac{12-3}{5-2}=\frac{9}{3}=3$  we often use the letter m for the ROC or slope.

Multiple Choice

What is the ROC for the following points?

 $\left(-1,\ 4\right),\ \left(7,\ 10\right)$  

•  $m\ =\ -\frac{3}{4}$  

•  $m=\frac{3}{4}$  

•  $m=\ -\frac{4}{3}$  

•  $m\ =\ \frac{4}{3}$  

Multiple Choice

What is the ROC for the following points?

 $\left(-2,\ -4\right),\ \left(1,\ 8\right)$  

•  $m=12$  

•  $m=-4$  

•  $m\ =\ -12$  

•  $m=4$  

Intersecting Functions

Often in math we want to know the Solution of a System of Equations. This means we want to know when two functions are equal to each other.

Intersecting Functions

To find the solution to a system of equations we set each equation equal to each other and then we find the point or points the two equations share in common.
Example: What is the solution to:

 $f\left(x\right)=x^2-3x+1\ and\ g\left(x\right)=x-2$  
Set each function equal to each other  $x^2-3x+1=x-2$  
Then get all terms on one side of the equation  $x^2-3x+1-x+2=0$  
Solve:  $x^2-4x+3=0$  
Factor:  $\left(x-3\right)\left(x-1\right)=0$  
Solutions for x are  $x=1\ or\ x\ =3$  

Intersecting Functions

Now the we know the inputs: $x=1\ or\ x=3$ 
We put them into either function to find the outputs (y)

 $g\left(1\right)\ =\left(1\right)-2\ or\ g\left(3\right)=\left(3\right)-2\$   $g\left(1\right)=-1\ or\ g\left(3\right)=2$  
Our solutions (point of intersection) are:  $\left(1,\ -1\right)\ or\ \left(3,\ 1\right)$  

Multiple Choice

Find the solution to the system of equations.

 $f\left(x\right)=x^2-4x+5\ and\ g\left(x\right)=3x\ -1$  

•  $\left(1,\ 2\right)\ and\ \left(6,\ 17\right)$  

•  $\left(-6,\ -19\right)\ and\ \left(-1,\ -4\right)$  

•  $\left(-6,\ -19\right)\ and\ \left(1,\ 2\right)$  

•  $\left(-1,\ -4\right)\ and\ \left(6,\ 17\right)$  

Quadratic Functions

A quadratic Function is any function with the highest power on a variable being two.

 $x^2$  
When we solve quadratics there are usually two answers but there can be one or no answers. 
The answers for a quadratic are when the graph crosses the x-axis (the zeros of the graph)

Quadratic Functions

  $x^2+4x-7=0$  

One way to find the zeros of a quadratic is to complete the square.

Here we want to get the constant on the other side of the equations:  $x^2+4x\ =\ 7$  
Then we complete the square by adding half of b squared to both sides  $\left(\frac{4}{2}\right)^2$  
 $x^2+4x+\left(2\right)^2=11$  
Now we have a perfect square of the right side  $\left(x+2\right)^2=11$  

Quadratic Functions

We will then take the square root of both sides (remember you get a positive root and a negative root)!

 $\sqrt{\left(x+2\right)^2}=\pm\sqrt{11}$  
 $x+2=\pm\sqrt{11}$  
 $x=-2\pm\sqrt{11}$  
So out solutions are  $-2+\sqrt{11\ }and\ -2-\sqrt{11}$  
Those are the values of x that make the equation equal to 0

Multiple Choice

Complete the Square

 $x^2+8x\ -4=0$  

•  $x=4\pm2\sqrt{5}$  

•  $x=4\pm4\sqrt{5}$  

•  $x=-4\pm2\sqrt{5}$  

•  $x=-4\pm4\sqrt{5}$  

The Quadratic Formula

 $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$  

We have a formula that can solve for the zeros of a quadratic it is called the Quadratic Formula.

a is the coefficient on the x², b is the coefficient on the x to the first degree and c is the constant number

The Quadratic Formula

To use the Quadratic Formula just plug in the numbers and solve!

 $x^2-7x+4=0$  
 $x=\frac{-\left(-7\right)\pm\sqrt{\left(-7\right)^2-4\left(1\right)\left(4\right)}}{2\left(1\right)}$  
 $x=\frac{7\pm\sqrt{49-16}}{2}$  

 $x=\frac{7\pm\sqrt{33}}{2}$  or  $x\ ≈\ 6.372\ and\ x\ ≈\ 0.628$  

Multiple Choice

Use the Quadratic Formula to solve:

 $x^2+5x-4=0$  

•  $x=\frac{5\pm\sqrt{41}}{2}$  

•  $x=\frac{\left(-5\pm\sqrt{41}\right)}{2}$  

•  $x=1\ and\ x\ =4$  

•  $x=-4\ and\ x=-1$