1. Quadratic Functions
By Libby Brown

2. THREE FORMS OF QUADRATIC EQUATIONS
Standard Form : f(x)=ax^2+bx+c
ex.) f(x)=3x^2+5x+2
Vertex Form: f(x)=a(x-h)^2+k
ex.) f(x)=4(x-3)^2+5
Intercept Form: f(x)=a(x-p)(x-q)
ex.) f(x)=6(x-2)(x-5)

3. TRICKS ABOUT QUADRATIC EQUATIONS
Standard Form
- if a>0, the graph opens up
- if a<0, the graph opens down
- if |a|>1, the graph is narrower than the parent function
- if |a|<1, the graph is wider than the parent function

4. TRICKS ABOUT QUADRATIC EQUATIONS (cont.)
Vertex Form
- a is the same as in Standard Form
- h makes the graph shift left and right (- h=left; +h=right)
- k makes the graph shift up and down (- k=down; +k=up)

5. TRICKS ABOUT QUADRATIC EQUATIONS (cont.)
Intercept Form
- a is the same as in Standard Form
- the y-intercept is apq

6. STANDARD FORM
The easiest way to solve an equation in standard form is to factor it and solve for x that way.
- Difference of Two Squares:
If both a and c in the equation are perfect squares, and b=0, you can use this method as a short cut for factoring.
ex.) 4x^2-16 can be factored as (2x-4)(2x+4), this way, the b term will cancel out when you foil the equation. Set both equations equal to zero, and you have your solutions.
- Perfect Square Trinomial:
If both a and c in the equation are perfect squares and b≠0, you might be able to use this method as a short cut for factoring. As long as the two values in the parentheses can be multiplied together and then doubled to equal the value of b in the original equation, you can use this method as a short cut for factoring.
ex.) x^2-4x+4 can be factored as (x-2)^2. Set the equation equal to zero and you will have your solution. However, if the equation was x^2-8x+4, you wouldn’t be able to factor this way.

7. STANDARD FORM (CONT.) Steps for Graphing in Standard Form
1.) The first step for graphing in standard form is to find the vertex of the parabola, (- b/2a, f(-b/2a)). To find the x value of the vertex, which is also the axis of symmetry, you use the equation x=-b/2a
ex.) If your equation is y=x^2-4x+6, you would plug in -4 for b and 1 for a. x= -(- 4)/2(1) = 4/2 = 2
2.) You would then plug the x value from step one back into the quadratic equation, and come up with the y coordinate of the vertex.
ex.) x=2; y=(2^2)-4(2)+6= 4-8+6= 2
3.) You now have the vertex of your parabola. The next step is to graph the y- intercept, which is the point (0,c), (0,6) in this case.
4.) Because points are evenly reflected over a parabola’s axis of symmetry, you can plot the point that is equidistant from the axis of symmetry as the y-intercept is. This point would be (2, 6).
5.) Connect and label the points.

8. VERTEX FORM
Because you can’t factor an equation in vertex form without converting it to standard form, the easiest way to solve an equation written in vertex form is to use the method of taking the square root of both sides. This only works when b=0 in the equation, which it always does when it is written in vertex form.
1.) First, you want to get (x-h)^2 alone, so you subtract k from both sides and then divide by a.
ex.) 2(x-5)^2-50=y
2(x-5)^2=50
(x-5)^2=25
2.) Next, you take the square root of both sides, remembering to use the ± sign.
ex.) x-5=±√25
x-5= ±5
3.) Finally, you get x alone by adding the remaining value to the opposite side, and you have your two solutions for x.
ex.) x= 5±5
x= 5-5=0
x= 5+5=10

9. VERTEX FORM (CONT.) Steps for Graphing in Vertex Form
1.) First, determine the vertex, (h,k). Remember that in the equation, h is being subtracted from x, so a positive h would be (x-h), and a negative h would be (x+h). Plot the vertex. You can also draw the axis of symmetry, because it is the x value of the vertex, which is h.
ex.) In the equation y=(x+2)^2+4, the vertex is (-2, 4), and the axis of symmetry is x=-2
2.) You can then find the coordinates for two other points on the graph – one to the right, and one to the left. A good trick to use is to go over by one, and up by a. This only works for one point to the left and one point to the right so don’t use this trick more than once!
3.) Connect the points and label.

10. INTERCEPT FORM
Because the solutions to a quadratic equation are where the parabola crosses the x-axis, or the x-intercepts, the solutions are in this form of equation for you. In the equation y=a(x-p)(x-q), p and q are the x-intercepts, and therefore they are the solutions to the equation. If you wanted to factor the equation, however, you would have to convert it to standard form first.

11. INTERCEPT FORM (CONT.) Steps for Graphing in Intercept Form
1.) Plot the x-intercepts of the equation, which are p and q. Don’t forget that p and q are being subtracted from x, so a positive value for p or q would be (x-p)(x-q), and a negative value for p or q would be (x+p)(x+q).
ex.) In the equation y=(x-4)(x+2), the x-intercepts are (4,0) and (-2,0).
2.) Find the coordinates of the vertex, (p+q/2, f(p+q/2)). You can also calculate the equation for the axis of symmetry, because it is the same as the x value of the vertex, p+q/2. After finding the x-coordinate of the vertex, plug it back into the original equation and solve for y. Plot the vertex
ex.) x= p+q/2= 4-2/2= 2 The vertex is (2, -8)
y= (2-4)(2+2)= (-2)(4)= -8
3.) Connect and label the points.

12. CONVERTING TO STANDARD FORM
Vertex to Standard
Multiply out and collect like terms
ex.) y=2(x+3)^ x^2+3x+3x+9= x^2+6x+9
y=2(x^2+6x+9)+2
y=2x^2+12x+18+2= 2x^2+12x+20
Intercept to Standard
ex.) y=2(x-2)(x-4) x^2-4x-2x+8=x^2-6x+8
y=2(x^2-6x+8)
y=2x^2-12x+16

13. THE END