03CE maths MCQ42

For a standard quadratic function y = ax^2 + bx + c, the axis of symmetry is given by x = - b/2a
Now a = - 1, b = a
so the axis of symmetry of the question is x = a/2
since the axis of symmetry of the graph is on the +ve side of the x - axis, that means a/2 > 0, therefore, a > 0.

To wy:

For y = ax² + bx + c
If a = -1, b = a,
then y = -x² - x + c
What do you want to say ?

Also, a = -1 and a > 0 is contradictory.

What a state of mess!

The answer is: (c) a > 0 and b < 0

y = -x² + ax + b
y = -(x² - ax) + b
y = -[x² - ax + (a/2)²] + (a/2)² + b
y = -[x - (a/2)]² + (a²/4) + b
Since -[x - (a/2)]² ≤ 0, y is maximum when x = a/2.
In the graph, when y is the maximum, x is positive.
Hence, a/2 is positive, and thus a > 0.

y = -x² + ax + b
When x = 0, y = b
The curve cuts y-axis at (0, b).
In the graph, the curve cuts y-axis at a point with negative y-coordinate.
Hence, b < 0