document.write( "Question 173836This question is from textbook Algebra 2
\n" ); document.write( ": My algebra 2 teacher has the equation $\"-2x%5E2=16x%2B28\"$ . I solved it by moving the -2x^2 to the right side (less work) to get $\"2x%5E2-16x-28\"$ . He solved it (more work) by moving 16x + 28 to the left side to get $\"-2x%5E2-16x-28\"$ . 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.\r
\n" ); document.write( "\n" ); document.write( "What is the rule? He says he always solves for a positve \"a\", but the book just says a cannot = 0.\r
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Algebra.Com's Answer #128705 by solver91311(24713) $\"\"$ $\"About$

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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). $\"-2x%5E2=16x%2B28\"$ is equivalent to $\"0=2x%5E2%2B16x%2B28\"$ . Furthermore, this is equivalent to $\"-2x%5E2-16x-28=0\"$ because it is perfectly valid to multiply both sides of an equation by -1 and completely swapping sides is perfectly valid as well.\r
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\n" ); document.write( "\n" ); document.write( "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.\r
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\n" ); document.write( "\n" ); document.write( "Now, if you want to look at the parabola associated with the quadratic equation, you can always write $\"f%28x%29=y=2x%5E2%2B16x%2B28\"$ which is, in fact a parabola whose vertex is a minimum. NOW if you multiply by a -1, you get: $\"-f%28x%29=-y=-2x%5E2-16x-28\"$ 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.\r
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\n" ); document.write( "\n" ); document.write( " $\"graph%28600%2C600%2C-10%2C10%2C-10%2C10%2C2x%5E2%2B16x%2B28%2C-2x%5E2-16x-28%29\"$ \r
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