Question

If the diagram in Fig. 2.17 shows the graph of the polynomial f(x) = ax² +bx+c, then:-
(a) a < 0, b < 0 and c > 0
(b) a < 0, b < 0 and c < 0
(c) a < 0, b > 0 and c > 0
(d) a < 0, b > 0 and c < 0

@BeBrainlyIndia​

Answer 1

Answer:

Please see the attachment

Answer 2

❒ Your Answer:

Here, we have a graph of a quadratic equation or polynomial ax² + bx + c, So the graph is parabolic curve$.$

• Clearly, The curve is facing downward, then a < 0

Let y = ax² + bx + c, that cuts the y axis at some point P which lies on line OY (Here). If, we put x = 0, then y = c

• So, coordinates of point P is (0,c). Hence, it lies on OY Cartesian Y axis, so c > 0.

Now, we can observe that the vertex of the parabola, ( -b/2a , -D/2a) lies on the second quadrant.

• Hence, b < 0

✒ So, Correct option is A.

❒ Example:

For your better understanding, here is an example of the above situation of graph formed by a quad. equation.

❇ Refer to the attachment...

Let y = f(x) or y = 3 - 2x - x² and now we will plot the graph of the equation and the required graph is in the attachment....

• x = -5, then y = -12
• x = -3, then y = 0
• x = -1, then y = 4
• x = 2, then y = -5
• x = 4, then y = -21

Here, highest point P(-1,4) is called the maximum point / maxima, and is the vertex of the polynomial.

⦿ Observations:

• The coefficient of x² in the quadratic eq. is -1 which is negative. So, parabola opens downwards.
• The parabola is factorised, I.e. (1 - x)(x + 3), So the parabola cuts the X axis at (1,0) and (-3,0)
• On comparing, this equation with ax² + bx + c, then a = -1, b = -2 and c = 3.

From here, we conclude that a < 0, b < 0 and c > 0

✒ Hence, verified graphically!

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