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If the diagram shows the graph of the polynomial $f x = a x 2 + b x + c$ , then

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a > 0, b < 0

and c

> 0

a < 0, b < 0

and

c < 0

a < 0, b > 0

and

c > 0

a < 0, b > 0

and

c < 0

answer is A.

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Detailed Solution

Given the polynomial

f x = a x 2 + b x + c

The graph of the quadratic polynomial

f x = a x 2 + b x + c

represents a parabola with its opening in the upward direction.
Thus,

a > 0

.
Vertex of the parabola lies in the fourth quadrant.
Substitute

x = 0

f (x) = a x 2 + b x + c,

⇒ f (x) = c

y = c .

So, the coordinates of the point in the graph are

(0, c) and so, c > 0

.
Thus,

a > 0, b < 0

and

c > 0

is possible.
If

a < 0, b < 0

and

c < 0,

then the curve is opening downward but the parabola is opening upward in the figure.
So,

a < 0, b < 0

and

c < 0

is not possible.
If

a < 0, b > 0

and

c > 0,

then the curve is opening downward but the parabola is opening upward in the figure.
So

a < 0, b > 0

and

c > 0

is not possible.
If the result is

a < 0, b > 0

and

c < 0,

then the curve is opening downward but the parabola is opening upward in the figure.
So,

a < 0, b > 0

and

c < 0

is not possible.
Therefore the only possible way is

a > 0, b < 0

and

c > 0

.
Hence the correct option is 1.

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