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.
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.
If you begin with a quadratic equation written as
y = ax^{2} + 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.
Factor out a and complete the square first:
y = ax^{2 }+ bx + c
y = a(x^{2}+(b/a)x+c/a) Factored out a.
y = a(x^{2 }+ 2(b/(2a))x + c/a) Prepared for completing square.
y = a(x^{2} + 2(b/(2a))x + (b/(2a))^{2} - (b/(2a))^{2} + c/a) Completed square.
y = a( (x + b/(2a))^{2} + (4ac - b^{2 })/(4a^{2}) ) Folded the square + combined terms.
y = a(x + b/(2a))^{2} + (4ac - b^{2})/(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 - b^{2})/(4a)
Comments
Ruth,
You hail from my old "stomping grounds" ! I'm a former Washingtonian...hail Redskins! Robert B's answer is a great one! I would add only one thing more...
Your vertex will have the abscissa (x-coordinate) of -b/2a , which comes from the formula he listed at the top. Be careful to use your signs wisely, as the "negative" will change the sign of that value. If you substitute that result back into the equation, then you will find the corresponding y-value to determine the ordered pair for the vertex. If you follow "Cramer" on CNBC and "Mad Money", he alludes to the parabola quite often to denote when a stock will "take off" based on the curve of the parabola. Any point to the right of the vertex will have a reflective y-value on the left. For example, if your vertex is at zero...on the y-axis, then the result from x=1 will be the same as for x= -1 on the other side. This also works for x=2 and -2. That makes YOUR work that much easier. By the way, GREAT job Robert. I will give you a thumbs up review. Contact either of us if you have any more questions.
Charles S.
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