# Quadratic Functions and Their Graphs

Graphing Basic Parabolic Functions.wmv

Tip: This isn't the place to ask a question because the teacher can't reply.

## Key Questions

• A quadratic function has the general form:
$y = a {x}^{2} + b x + c$

(where $a , b \mathmr{and} c$ are real numbers) and is represented graphically by a curve called PARABOLA that has a shape of a downwards or upwards U.
The main features of this curve are:
1) Concavity: up or down. This depends upon the sign of the real number $a$:

2) Vertex. The vertex is the highes or lowest point of the parÃ¡bola.
the coordinates of this point are:
$x = - \frac{b}{2 a}$ and $y = - \frac{\Delta}{4 a}$
Where $\Delta = {b}^{2} - 4 a c$

3) point of intercept with the y axis. This is the point where the parÃ¡bola crosses the y axis and has coordinates: $\left(0 , c\right)$

4) Possible points of intercept with the x axis (there also can be none). These are the points where the parÃ¡bola crosses the x axis.
They are obtained by putting y=0 and solving for x the 2nd degree equation: $a {x}^{2} + b x + c = 0$, which will give the x coordinates of these points (2 solutions)
Depending on the discriminant $\Delta = {b}^{2} - 4 a c$ if it is <0 the parÃ¡bola does not cross the x axis.

• There are two forms a quadratic function could be written in: standard or vertex form. Here are the following ways you can determine the vertex and direction dependent on the form:

Standard Form ($f \left(x\right) = a {x}^{2} + b x + c$)
1. Direction of the parabola can be determined by the value of a. If a is positive, then the parabola faces up (making a u shaped). If a is negative, then the parabola faces down (upside down u).
2. Vertex can be found by $x = - \frac{b}{2 a}$ and then plugging in that value to find y.
Here is an example:
$y = - 2 {x}^{2} + 4 x - 3$, Faces downward since a = -2.
to find the vertex: $x = \frac{- 4}{2 \left(- 2\right)} = \frac{- 4}{- 4} = 1$
then plug that value into the equation$y = - 2 {x}^{2} + 4 x - 3$
$y = - 2 {\left(1\right)}^{2} + 4 \left(1\right) - 3$
$y = - 2 \left(1\right) + 4 \left(1\right) - 3$
$y = - 2 + 4 - 3$
$y = - 1$
Vertex is (1,-1)

Vertex Form ($y = a {\left(x - h\right)}^{2} + k$)
1. Direction of the parabola can be determined by the value of a. If a is positive, then the parabola faces up (making a u shaped). If a is negative, then the parabola faces down (upside down u).
2. Vertex is (h,k).
Here is an example:
$y = - 3 \left(x - 2\right) + 6$ Faces down since a = -3 and the vertex is (2, 6).

• A quadratic function is one of general form:
$y = a {x}^{2} + b x + c$

where a, b and c are real numbers.

This function can be plotted giving a PARABOLA (a curve in the shape of an upward or downward U)

To find the x intercepts you must put y=0; in this way you fix at zero the coordinate y of the points you are seeking.
You are left with finding the coordinate x of the points.
If y=0 you are left with: $0 = a {x}^{2} + b x + c$ which is a second degree equation.
By solving this equation you'll find two values of x (x1 and x2) that together with y=0 will give you the intercepts:
intercept 1: (x1 , 0)
intercept 2: (x2 , 0)

Remember that a second degree equation can also have solutions:
- coincident (the intercept is the VERTEX of the parabola)
- imaginary (The parabola does not cross the x axis)
Depending upon the discriminant of the equation.

• The graphs of quadratic functions are called parabolas, and they look like the letter "U" right side up or up side down.

I hope that this was helpful.

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