Create a unique parabola in the pattern f(x)= ax^2 + bx + c. Describe the direction of the parabola and determine the y intercept, zeros, vertex, and the equation for the axis of symmetry

Answer 1

The problem asks for standard form. Since we need the zeros (where Y = 0), let's start out in intercept form and work back.

Lets try:

(x+2)(x+4)

The zeros are (-2,-4) (Y will equal 0)

The x coordinate of the vertex is given by the sum of the zeros over 2. This is because the x coordinate of the vertex is the middle of any two points on the parabola with the same Y value. (axis of symmetry) The zeros are actually two separate x coordinates where Y=0.

(-2+-4)/2 = -3. This is the x coordinate of the vertex. Plug this back into the formula. (-3+2)(-3+4) = -1*1 = 1

So the vertex is (-3,-1)

Axis of symmetry is the line with goes vertically through the vertex. That's x =-3

Now foil to put in standard form. = x²+ 6x + 8. The Y intercept is when x = 0, so Y intercept = 8

Finally, the direction of the parabola always follows the sign on A. (in all three forms actually)

A is positive so the parabola is up (happy face)

For the graph, input into desmos

Hope this helps!

Note: you can also figure out the vertex from standard from with the equation -b/2a = --6/2 = -3