Question

both sides of an equation have quadratic expressions. the graph of one quadratic opens upward. the graph of the other opens downward. what are the possible number of intersections for this system? 0 1 2 infinite done

Explanation:

Step1: Set up the system

Let upward quadratic: $y = ax^2+bx+c$ ($a>0$), downward quadratic: $y = dx^2+ex+f$ ($d<0$). Set equal: $ax^2+bx+c = dx^2+ex+f$

Step2: Simplify to single quadratic

Rearrange: $(a-d)x^2+(b-e)x+(c-f)=0$. Since $a-d>0$ (sum of positive and positive, as $d<0$), this is a quadratic opening upward.

Step3: Analyze discriminant

For quadratic $Ax^2+Bx+C=0$, discriminant $\Delta=B^2-4AC$.

• $\Delta<0$: 0 real solutions (0 intersections)
• $\Delta=0$: 1 real solution (1 intersection)
• $\Delta>0$: 2 real solutions (2 intersections)

Infinite intersections only if quadratics are identical, impossible here (one opens up, one down).

Answer:

0, 1, 2