Answer:
Option (1)
Step-by-step explanation:
Given quadratic equation in this question is,
y < -x²+ 4x + 5
Now we will convert this standard quadratic equation into vertex form,
y < -(x² - 4x) + 5
y < -[x² - 2(2)x + 2²] + 5
y < -(x - 2)²+ 5
This equation is in the form of y < a(x - h)² + k
where (h, k) is the vertex of the parabola.
Therefore, y < -(x - 2)²+ 5 will show the parabola with properties as,
1). Parabola having vertex at (2, 5).
2). Coefficient 'a' is negative, so parabola will open downwards.
3). In the inequality notation of less than, (sign < ) will show the solution area inside the parabola.
[If an inequality has a sign of greater than, solution area will be outside the parabola.]
Therefore, Option (1) will be the answer.
