The graph of y = ax 2 + bx + c is shown below. Determine the solution set of 0 = ax 2 + bx + c.
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Answer:
look below
Step-by-step explanation:
The vertex will be found at (4,2) and we can write that :
(y - 2) = (a)(x - 4)^2 and since (8,6) is on the curve, we can solve for a
(6 -2) = (a) (8 - 4)^2 simplify
4 = (a)(4)^2
4 = (a)16 → a = 4/16 = 1/4
Since the x coordinate of the vertex is given by -b/ (2a) we have that -b/[2 (1/4)] = 4 → -b / (1/2) = 4 → -b = 2 → b = -2
And using the fact that (4,2) is on the graph, we can find c, thusly :
y = ax^2 + bx + c ......so......
2 = (1/4)(4)^2 -2(4) + c
2 = 4 - 8 + c
2 = -4 + c
c = 6
Then a*b*c = (1/4) (-2) (6) = (1/4)(-12) = -3
The vertex at (4,2) can be found by
(y - 2) = (a)(x - 4)^2
∵ (8,6) is lying on the curve
So, (6 -2) = (a) (8 - 4)^2
by simplify, we get
4 = (a)(4)^2
4 = (a)16
a = 4/16 = 1/4
Since the coordinate of x-axies of the vertex is given by -b/(2a) we have
b/[2 (1/4)] = 4
-b / (1/2) = 4
-b = 2
b = -2
by using the point that (4,2) is on the graph, we can find c
y = ax^2 + bx + c
2 = (1/4)(4)^2 -2(4) + c
2 = 4 - 8 + c
2 = -4 + c
c = 6
∴ a*b*c
= (1/4) (-2) (6)
= (1/4)(-12)
= -3
Diagrams of direct conditions are particularly successful for addressing connections between things that change at a steady rate, and they frequently improve than words or numerical conditions alone.
The diagram of a quadratic capability is known as a parabola and has a bent shape. One of the primary concerns of a parabola is its vertex. It is the most noteworthy or the absolute bottom on the diagram.
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