Multiple choice

The graphs of the two equations $y = ax^2+bx+c$ and $y =$ $Ax^2+Bx+C$, such that $a$ and $A$ have different signs and that the quantities $b^2 - 4ac$ and $B^2-4AC$ are both negative,

  1. intersect at two points

  2.  intersect at one point

  3. do not intersect

  4. none of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The condition b^2 - 4ac < 0 implies the parabola y = ax^2+bx+c never crosses the x-axis. Since a and A have different signs, one parabola opens upward and the other downward. Because they never cross the x-axis, they are entirely on opposite sides of the x-axis, meaning they cannot intersect.

AI explanation

Because the discriminant b squared minus 4 times a times c is negative for both parabolas, their graphs do not intersect the x-axis and they each have only real y-intercepts. Since the leading coefficients a and A have different signs, one parabola opens entirely upwards while the other opens entirely downwards. A downward opening parabola with no real roots lies completely below the x-axis, while an upward opening one lies completely above it. Consequently, their graphs do not intersect.

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