Question

The graphs of $y=a{x}^2+bx+c$ are given in Figure. Identify the signs of a, b and c in each of the following:

Answer 1

(i) We observe that $y=a{x}^2+bx+c$ represents a parabola opening downwards.

Therefore, a < 0. We also observe that the vertex of the parabola is in first quadrant.

$\therefore$ $-\frac{b}{2a}>0\Rightarrow -b<0\Rightarrow b>0$ $[\because a<0]$

Parabola $y=a{x}^2+bx+c$ cuts Y-axis at P. On Y-axis, we have x = 0.Putting x = 0 in $y=a{x}^2+bx+c$, we get y = c.So, the coordinates of P are (0, c). As P lies on the positive direction of Y-axis. Therefore, c> 0.Hence, a < 0, b > 0 and c> 0.

(ii) We find that $y=a{x}^2+bx+c$ represents a parabola opening upwards. Therefore, a > 0. The vertex of the parabola is in fourth quadrant.

$\therefore$ $-\frac{b}{2a}>0\Rightarrow -b<0\Rightarrow b>0$ $[\because a<0]$

Parabola $y=a{x}^2+bx+c$ cuts Y-axis at P and on Y-axis. We have x = 0. Therefore, on putting x = 0 in $y=a{x}^2+bx+c$, we get y = c.So, the coordinates of P are (0, c). As P lies on OY'. Therefore, c < 0.\Hence, a > 0, & < 0 and c < 0.

(iii) Clearly, $y=a{x}^2+bx+c$ represents a parabola opening upwards.Therefore, a > 0. The vertex of the parabola lies on OX.

$\therefore$ $-\frac{b}{2a}>0\Rightarrow -b<0\Rightarrow b>0$ $[\because a<0]$

The parabola $y=a{x}^2+bx+c$ cuts Y-axis at P which lies on OY.

Putting x = 0 in $y=a{x}^2+bx+c$, we get y = c. So, the coordinates of P are (0, c). Clearly, P lies on OY.

$\therefore$ c> 0. Hence, a > 0, b < 0, and c > 0.

(iv) The parabola $y=a{x}^2+bx+c$ opens downwards. Therefore, a < 0.

The vertex $(\frac{-b}{2a},\frac{-D}{4a})$ of the parabola is on OX'.

$\therefore$ $\frac{-b}{2a}<0\Rightarrow b<0$ $[\because a<0]$

Parabola $y=a{x}^2+bx+c$ cuts Y-axis at P (0, c) which lies on OY'.

Therefore, c < 0. Hence, a < 0, & < 0 and c < 0.

(v) We notice that the parabola $y=a{x}^2+bx+c$ opens upwards. Therefore, a > 0. Vertex

$(\frac{-b}{2a},\frac{-D}{4a})$ of the parabola lies in the first quadrant.

$\therefore$ $\frac{-b}{2a}>0\Rightarrow \frac{b}{2a}<0\Rightarrow b<0$ $[\because a>0]$

As P (0, c) lies on OY. Therefore, c > 0. Hence, a > 0, b < 0 and c > 0.

(vi) Clearly, a < 0. As P $(\frac{-b}{2a},\frac{-D}{4a})$ lies in the fourth quadrant.$\therefore$ $\frac{-b}{2a}>0\Rightarrow \frac{b}{2a}<0\Rightarrow b<0[\because a>0]$

As P (0, c) lies on OY'. Therefore, c < 0. Hence, a < 0, b > 0 and c < 0.