SOLUTION: Currently graphing parabolas, please explain how to put y=x^2+2x into standard form.
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Question 884071: Currently graphing parabolas, please explain how to put y=x^2+2x into standard form.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
if the solution you are looking for is (x+1)^2 - 1, then that is derived in the following manner:
(x+1)^2 = x^2 + 2x + 1
subtract 1 from both sides of that equation and you get:
(x+1)^2 - 1 = x^2 + 2x
those 2 forms of the equation are equivalent.
completing the square is part of the completing the square method.
let's take your equation of y = x^2 + 2x.
since this is in standard form of y = ax^2 + bx + c, then a = 1 and b = 2 and c = 0.
we factor this by taking half the coefficient of the x term and then forming the equation of:
y = (x+1)^2 - 1
as shown above, this is equivalent to y = x^2 + 2x.
we now set the y equal to 0 to get:
(x+1)^2 - 1 = 0
we then add 1 to both sides of this equation to get:
(x+1)^2 = 1
we then take the square root of both sides of this equation to get:
x+1 = +/- sqrt(1)
we then subtract 1 from both sides of this equation to get:
x = -1 +/- sqrt(1) which results in:
x = 0 and x = -2
those should be the roots of the equation which are the solutions of the equation.
when x = 0, x^2 + 2x = 0 which is correct.
when x = -2, x^2 + 2x = 4 - 4 = 0 which is also correct.
the roots are where the graph of the equation crosses the x-axis which is at the points of x = 0 and x = -2
the graph of the equation is shown below:
the completing the square method is not used very much since it is a little bit complex and cumbersome to use, but it does work.
mostly you will solve quadratic equations by factoring or by using the quadratic formula.
some references that might be helpful for you are:
http://www.regentsprep.org/Regents/math/algtrig/ATE3/QuadLesson.htm
http://tutorial.math.lamar.edu/Classes/Alg/SolveQuadraticEqnsII.aspx
http://www.purplemath.com/modules/sqrquad.htm
http://www.themathpage.com/aprecalc/complete-the-square.htm
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