how do you determine the axis of symmetry, vertex, and two points?

Question

I'm trying to learn about quadratic functions in standard form: conceptual enrichment.

 

Determine the axis of symmetry, vertex, and at least two points (x, y). The graph the quadratic equation.

Answer 1

If you begin with a quadratic equation written as

y = ax² + bx + c,

it will be best to rewrite it another way.

By "completing the square," you should be able to write it as

y = a(x - h)² + k.

Once in this form, everything is a little easier to find.

The vertex is located at (h,k).  The axis of symmetry is the vertical line through the vertex, or x = h.

You can find two other points by choosing two values for x (other than h), and find the corresponding y-values.  This will give you two more ordered pairs to assist you in graphing the equation.  Choosing one value below h, and one above it will give you points on both sides of the vertex.

Hope this helps.  If you need further assistance, I or another tutor will be happy to help.

Answer 2

Factor out a and complete the square first:

y = ax² + bx + c

y = a(x²+(b/a)x+c/a) Factored out a.

y = a(x² + 2(b/(2a))x + c/a) Prepared for completing square.

y = a(x² + 2(b/(2a))x + (b/(2a))² - (b/(2a))² + c/a) Completed square.

y = a( (x + b/(2a))²  + (4ac - b² )/(4a²) ) Folded the square + combined terms.

y = a(x + b/(2a))² + (4ac - b²)/(4a) distributed a.

So the equation of the axis of symmetry is x = -b/(2a), where the square expression vanishes.

The vertex is on this axis and has y-coordinate y = (4ac - b²)/(4a)