Question 173836This question is from textbook Algebra 2
: My algebra 2 teacher has the equation  . I solved it by moving the -2x^2 to the right side (less work) to get  . He solved it (more work) by moving 16x + 28 to the left side to get  . We both get the same roots and the same axis of symmetry, but different vertices. I get a minimum, my parabola is "upright" while he gets a maximum because his is upside down.
What is the rule? He says he always solves for a positve "a", but the book just says a cannot = 0.
Help!
This question is from textbook Algebra 2
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website! Your problem is that you are beginning with a quadratic equation and confusing that with a function whose graph is in the shape of a parabola. Furthermore, you have made a sign error when you rearranged the equation by putting everything on the right (although I suspect this was simply a typo when you posted this problem, otherwise you would not have obtained the same pair of roots as your instructor did). is equivalent to  . Furthermore, this is equivalent to  because it is perfectly valid to multiply both sides of an equation by -1 and completely swapping sides is perfectly valid as well.
The simple fact is that all of the above equations have a solution set that consists of the two roots of the equation. They do NOT represent a parabola which is a curve that consists of an infinite number of points.
Now, if you want to look at the parabola associated with the quadratic equation, you can always write  which is, in fact a parabola whose vertex is a minimum. NOW if you multiply by a -1, you get:  and, as you might expect, your graph will be a parabola with the vertex as a maximum. The two functions do, in fact, have the same x-intercepts as you would expect from the results of solving the original quadratic equation.
|
|
|
