By: Suhasee

Differences Between Quadratic and Linear Functions

How can you tell whether an equation is a quadratic relation or a linear relation, just by looking at it?
By looking at a table of values, you can easily see whether an equation is a quadratic relation or a linear relation. This is because, by doing first and second difference, you can determine the type of relation. If the first difference is constant, then it is a linear relation. If the first difference isn't constant, but the second difference is, then it is a quadratic relation.

What is a Parabola?

The shape made by the Quadratic function is also known as a Parabola.
A parabola has many different components:
• Vertex (x,y) -- the point where the parabola changes direction.
• The Optimal Value (OV)-- the y-value of the vertex.
• The Axis of Symmetry (AOS)-- the x-value of the vertex, and passes through the x-value.
• Zero--is/are the X-Intercept(s) of the parabola.
• Y-Intercepts-- is the point where the parabola touched the y-axis.

Summary of Quadratics in Vertex Form

Vertex Form is one of the ways to write a quadratic function. Just by looking at the equation, you could identify the transformations of the base graph, the vertex, the direction of opening.
Written as: y= a(x-h)²+k ​

Different Translations in y= a(x-h)²+k ​

The value of (a), is either a vertical stretch or compression, by the factor of (a). It would be a stretch, if the number was a whole number. If the value was either a fraction or a decimal, then it would be a compression. If the value is negative, then it means that the parabola was reflected over the x-axis.

The value (-h) that is inside the bracket, is the horizontal translation. The value of h, moves the parabola either left or right depending or it's sign. (If the sign is negative, it moves right, if it is positive move left)--Opposite operation.

The value of (k), is the vertical translation. It moves the parabola either up or down depending on the sign. (If the sign is negative, it moves down and if it is positive, it moves up).
Finding Y-Intercept and Zero

To solve a quadratic equation, means to find the x-intercept or the roots. There could be many possible solutions, however there would only be one y-intercept.

To find the X-Intercept(s) or the Y-Intercept, set the opposite variable to 0.

Mapping Notation

1. Create a table of values using the base graph (y=x²)

2. Then you would look at the equation and substitute the a, h & k values into the mapping formula (x+h),(ay+k).

3. Use the mapping formula to make another table of values for the equation. Apply the first part of the formula (x+h) to the x values, and the second part (ay+k) to the y values, of the base function.

4. Using the values for x and y, from the new table of values, graph the point on a set of axis, draw the parabola.

Step Pattern

1. Find the vertex from the equation (it would be the (h,k) values)

2. You would then plot the points of the vertex.

3. You would look at the a value in the equation and then you would multiply it by 1, 3 ,5 etc...

4. Whatever value that you get you would start from the vertex and go one to the right and however many units up. You would keep on going until you find the co-ordinates of the parabola.

Below is the video that can be used to further explain the step pattern.

Word Problems Using Vertex Form

Example Word Problem:

A flare is released into the air following the path, h=-5(t-6)² +182, where h is the height (m), and t is time (s).

a) What is the maximum height?

The maximum height of the flare is 182m.

b) What is the initial height?

You would have to find the y-intercept and in order to do that you would have to substitute t as 0 in order to find the height.

h=-5(0-6)² +182

h=-5(36) + 182

h=-180+182

h=2

Therefore, the initial height of the flare is 2m.

c) How long was the flare in the air?

You would have to find the x-intercept and in order to do that you would have to substitute h as 0 in order to find the time that the flare was in the air.

h=-5(t-6)² +182

0==-5(t-6)² +182

0-182=-5(t-6)²

-182 /5= (t-6)²

√36.4=t-6

6.03=t-6

6.03+6=t

12.03=t

Therefore, the flare in the sky was 12.03 seconds.