y = -2x² + 8x + 3
Factor the coefficient of the x² term out of the x-terms on the right
[Do not factor out the x, but only the numerical coefficient]
y = -2(x² + 4x) + 3
Multiply the coefficient of x by 1/2, then square what you get.
4(1/2) = 2, then 2² = 4
Add and then subtract that number inside the parentheses:
[This just amounts to adding 0 which does not change the
value]
y = -2(x² + 4x + 4 - 4) + 3
Change the parentheses to brackets so you can insert parentheses
around the first three terms:
y = -2[(x² + 4x + 4) - 4] + 3
Factor the trinomial in the parentheses which should turn
out to be the square of a binomial:
y = -2[(x + 2)(x + 2) - 4] + 3
y = -2[(x + 2)² - 4] + 3
Remove the brackets by multiplying the -2 by each of
the terms inside the bracket, keeping the (x + 2)² intact:
y = -2(x + 2)² + 8 + 3
y = -2(x + 2)² + 11
That's it.
Compare to
y = a(x - h)² + k
and you will see that the vertex is (-2,11)
-------------------------------------------
y = 3x² - 6x + 5
Factor the coefficient of the x² term out of the x-terms on the right
[Do not factor out the x, but only the numerical coefficient]
y = 3(x² - 2x) + 5
Multiply the coefficient of x by 1/2, then square what you get.
-2(1/2) = -1, then (-1)² = 1
Add and then subtract that number inside the parentheses:
[This just amounts to adding 0 which does not change the
value]
y = 3(x² - 2x + 1 - 1) + 5
Change the parentheses to brackets so you can insert parentheses
around the first three terms:
y = 3[(x² - 2x + 1) - 1] + 5
Factor the trinomial in the parentheses which should turn
out to be the square of a binomial:
y = 3[(x - 1)(x - 1) - 1] + 5
y = 3[(x - 1)² - 1] + 5
Remove the brackets by multiplying the 3 by each of
the terms inside the bracket, keeping the (x + 2)² intact:
y = 3(x - 1)² - 3 + 5
y = 3(x - 1)² + 2
That's it.
Compare to
y = a(x - h)² + k
and you will see that the vertex is (1,2)
Edwin
