SOLUTION: determine the quadratic function f whose vertex is (2,-4) and passes through (1,-2)
f(x)=?
answer question in form of f(x)=ax^2+bx+c
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Question 638927: determine the quadratic function f whose vertex is (2,-4) and passes through (1,-2)
f(x)=?
answer question in form of f(x)=ax^2+bx+c
Found 2 solutions by reviewermath, DrBeeee:
Answer by reviewermath(1029) (Show Source): You can put this solution on YOUR website!
In , the vertex is (h,k).
The vertex is (2, -4) therefore h = 2 and k = -4.
, substitute (1,-2) and solve for "a".
.
Answer:
Answer by DrBeeee(684) (Show Source): You can put this solution on YOUR website!
General equation of a quadratic is
(1) f(x) = ax^2 + bx +c.
The graph of this quadratic is called a parabola. It is a curve shaped like the letter "U" or an upside down "U" depending on the sign of "a", the coefficient of x^2. If a>0 the parabola opens "up", like U. If a<0, then parabola opens "down". A useful formula is the value of x around which the parabola is symmetrical. This value is given by
(2) x = -b/(2a)
The value of x in (2) is also value of the x coordinate of the parabola's vertex, the point where the graph of f(x) hits a maximum or minimum.
For this problem we are given two points through which the parabola passes. The point
(3) (1,-2)
and and the vertex
(4) (2,-4).
Using (2) and the value of x in the vertex (4) yields
2 = -b/(2a) or
(5) b = -4a
Substituting b into (1) yields
(6) y = ax^2 -4ax + c
Now substitute the point (1,-2) of (3) into (6) yields
(7) -2 = a*(1)^2 - 4a*(1) + c, or simplifying
(8) c = -3a - 2
Now substitute the vertex (2,-4) of (4) into (6) yields
(9) -4 = a*(2)^2 -4a*(2) + c, or simplifying
(10) c = 4a - 4
Equating (8) and (10), and simplifying yields
(11) a = 2
Put a = 2 into (5) yields
(12) b = -8
Put a = 2 into (8) and simplifying yields
(13) c = 4
Check our answer in (7)
Is [-2 = 2*1 - 4*2*1 + 4]?
Is [-2 = 2 - 8 + 4]?
Is [-2 = -2]? Yes
Check our answer in (9)
Is [-4 = 2*2^2 -4*2*2 +4]?
Is [-4 = 8 - 16 +4]?
Is [-4 = -4]? Yes
Answer: f(x) = 2x^2 -8x +4
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