Given figure shows the graph of the polynomial f(x) = ax + bx + c ax2+ bx p o x (a) a b <O and c >O (c) n b and (b) a b <Oand c (d) a b>Oand c

Question

Answer 1

∵ vertex of parabola (\(\frac{-b}{2a}, \frac{-D}{4a}\)) lies in 4^th quadrant in which y coordinate is negative.

∴ \(\frac{-D}{4a} < 0\)

but D = \(\sqrt{b^2 - 4ac} > 0\)

∴ -1/a < 0 \(\Rightarrow\) 1/a > 0 \(\Rightarrow\) a > 0

In 4th quadrant x - coordinate in always positive.

∴ \(\frac{-b}{2a} > 0\)

put a > 0

∴ -b > 0

\(\Rightarrow\) b < 0

∵ graph of f(x) cuts y-axis at positive y-axis.

∴ y-coordinate of intersection point is positive.

∴ c > 0 (∵ f(0) = c and (0, c) is intersection point of curve f(x) and y - axis)

∴ a > 0, b < 0 and c > 0

option (b) is correct.