Welcome to Graphing Parabolas
By
Sahand Reiisieh

Parabola: The set of all points equidistant from a point called the focus and a line called the directrix. The focus lies on the axis of symmetry, and the directrix is perpendicular to the axis of symmetry.

Note: The graph of a quadratic formula y = ax2 + bx + c is a parabola

Keys to Success:

. Find the axis of symmetry. The axis of symmetry is an imaginary dashed line that runs through the vertex equally dividing the parabola in half. The formula for finding the line of symmetry is x = -b/2a

. Find the X and Y intercepts. These are the EASIEST points to find to help find the point on your parabola. These points can be found by setting X to zero and solving for Y, and vice versa.

. A positive ax2 gives you a parabola in the shape of a smiley face. Any negative coefficient before the ax will give you a parabola shape of a sad of frowning face ( an upside down U).

. To find the vertex simply take the value of X found by computing the axis of symmetry. Insert that value for all X's in the quadratic equation and solve for Y. The values of (X,Y) will be your vertex in which the axis of symmetry runs through.

Example Problem:

Graph y=2x2-8x+6

Step 1: Find the line of symmetry
. x=-b/2a
. x= -8/4
. x = 2

Step 2: Plug the value of x=2 into the equation.
. y= 2(2)2-8(2) + 6
. y= -2

Step 3: Plot the Vertex
. (2,-2)

Step 4: Find more points!
. Set x to zero to find the Y intercept.
. y= 2(0)2-8(0) + 6
. y= (0, 6)

Step 5: Use Symmetry
. Use symmetry to plot more points by referring to the axis of symmetry.
. Since (0,6) is -2 spaces away from the axis simply move over +2 spaces to the right of the axis to find a second point (4,6)

Now YOU try =)

Practice problems

Graph the equations. Find axis of symmetry and vertex
1) y= 2x2-12x+19
2) y= -3x2+0x+5
3) y=-1/6x2-x-3