Question

If the given figure shows the graph of polynomial \( y = ax^2 + bx + c \), then:

• A. \( a<0 \)
• B. \( b^2<4ac \)
• C. \( c>0 \)
• D. \( a \text{ and } b \text{ are of same sign} \)

Hint:

For a quadratic graph, if it opens downward, the coefficient \( a \) of \( x^2 \) is negative. Always check the opening direction of the parabola to infer the sign of \( a \).

Answer 1

The Correct Option is A

The graph shown is of a quadratic polynomial \( y = ax^2 + bx + c \). It is a parabola opening downward. For any quadratic equation, the direction in which the parabola opens depends on the sign of the leading coefficient \( a \):
- If \( a>0 \), the parabola opens upwards.
- If \( a<0 \), the parabola opens downwards.
In the given graph, since the parabola opens downwards, it is clear that:
\[ a<0 \] Let's analyze other options:
- Option (B): \( b^2<4ac \) implies the roots are imaginary, but the graph intersects the x-axis at two points, so roots are real and distinct. This is incorrect.
- Option (C): \( c>0 \) implies the y-intercept is positive. However, the graph intersects the y-axis below the x-axis, meaning \( c<0 \). This is also incorrect.
- Option (D): The sign of \( b \) cannot be determined from the graph alone without knowing the axis of symmetry. So this option is not necessarily true.
Hence, only option (A) is correct.