+42 Using the completing-the-square method, find the vertex of the function f(x) = –2x2 + 12x + 5 and indicate whether it is a minimum or a maximum and at what point.
A. Maximum at (–3, 5)
B. Minimum at (–3, 5)
C. Maximum at (3, 23)
D. Minimum at (3, 23)
+129852 f(x) = –2x2 + 12x + 5 factor out the "-2" on the right on the first two terms
f(x) = -2 (x^2 - 6x) + 5 take 1/2 of the coefficient on x (i.e., "-6"→ ( -3)^2 → 9)......square it and add an subtract it within the parentheses
f(x) = -2(x^2 - 6x + 9 - 9) + 5 simplify
f(x) -2 ( x ^2 - 6x + 9) + 18 + 5 factor the expression in the parenthese
f(x) = -2 (x - 3)^2 + 23
The vertex is at (3, 23)......since there is a "negative" in front of the first term, this is a maximum
Here's the graph : https://www.desmos.com/calculator/smeqago7cm
+129852 f(x) = –2x2 + 12x + 5 factor out the "-2" on the right on the first two terms
f(x) = -2 (x^2 - 6x) + 5 take 1/2 of the coefficient on x (i.e., "-6"→ ( -3)^2 → 9)......square it and add an subtract it within the parentheses
f(x) = -2(x^2 - 6x + 9 - 9) + 5 simplify
f(x) -2 ( x ^2 - 6x + 9) + 18 + 5 factor the expression in the parenthese
f(x) = -2 (x - 3)^2 + 23
The vertex is at (3, 23)......since there is a "negative" in front of the first term, this is a maximum
Here's the graph : https://www.desmos.com/calculator/smeqago7cm
