answer for math questions about parabolas
x^{2 }- 2x + 3 = (x - 1)^{2} + 2
= (x - 1)^{2} - (-2)
= (x - 1 + sqrt(-2)) (x - 1 - sqrt(-2))
= (x - 1 + sqrt(2) * i) (x - 1 - sqrt(2) * i)
answer for math questions about parabolas
x^{2 }- 2x + 3 = (x - 1)^{2} + 2
= (x - 1)^{2} - (-2)
= (x - 1 + sqrt(-2)) (x - 1 - sqrt(-2))
= (x - 1 + sqrt(2) * i) (x - 1 - sqrt(2) * i)
Parabolas are of the form ax^{2}+bx+c, where a,v and c are the coefficients (numbers) in front of the variable. Notice that all parabolas are of degree 2 (that means the highest power of the variable is 2).
There are three major methods of solving for the roots of a parabola. (roots being the places where the y value will be zero).
1. factor the tri-nomial (three terms) into two binomials (two terms).
2. complete the square (see below)
3. or use the quadratic formula -b + sqrt(b^{2} - 4ac) then divide the whole thing by 2a
for the parabola stated above: y =
x^{2} - 2x + 3 factoing will not be the best way.
completing the square is done as follows:
set the equation = 0. 0 = x^{2} -2x + 3
take the last term (3) and subtract it from both sides which gives you x^{2 }- 2x^{ = }-3
take the coefficient of the 'x' term which is -2, take half of it (-1) square it (1)
add this term (1) to both sides which gives you x^{2} - 2x +1 = -2
NOTICE: you have made a perfect square of the left side which is (x-1)^{2} so....
(x-1)^{2} =-2. Now take the sqrt of both sides which gives you x-1 =+sqrt(-2)
so..... the roots are complex! not real. they are x = 1-i*sqrt(2) and 1+i*sqrt(2)
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NOTE: The vertex or (maximum/minimum point) of the parabola ca be gotten by using the co-ordinate pair (-b/2a,f(-b/2a))
((-2/2),6) or ((-1,6)
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Using the quadratic formula we get -(-2) + sqrt(-2^{2}-4(a)(3) all divided by (2*1)
which gives you 2 + sqrt(4 - 12) all divided by 2
which gives you 2 + sqrt(-8) all divided by 2, which yields 2 + 2i* sqrt(2) alld divided by 2
whihc is 1 + i*sqrt(2) and 1 - i*sqrt(2)
same answer as completing the square!