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f(x) = 2x² + 12x + 15
We want to get it to the standard form
f(x) = a(x - h)² + k
f(x) = 2x² + 12x + 15
Factor 2 out of the first two terms:
f(x) = 2[x² + 6] + 15
Skip a space on the right of the bracketed expression
f(x) = 2[x² + 6x ] + 15
To the side, complete the square:
6(1/2) = 3 <--half of the coefficient of x
3² = 9 <--square that result
Add and subtract the value inside the bracket:
f(x) = 2[x² + 6x + 9 - 9] + 15
Factor the first three terms inside the bracket:
f(x) = 2[(x+3)(x+3) - 9] + 15
Write the factorization as the square of (x+3)
f(x) = 2[(x+3)² - 9] + 15
Distribute the 2 inside the bracket leaving the (x+3)² intact:
f(x) = 2(x+3)² - 18 + 15
Combine like terms:
f(x) = 2(x+3)² - 3
Compare to
f(x) = a(x-h)² + k
The vertex is (h,k) = (-3,-3)
Get a couple of points on each side of
the vertex (-3,-3)
x | y = f(x)
--|---------
-5 | 5
-4 | -1
-3 | -3 <-- vertex
-2 | -1
-1 | 5
Edwin
