Question

The parabola y=-x^2+12x-11 has an axis of symmetry of x=6. which of the following represents its range?

1) y>-11
2)y<25
3)y<6
4)y>10

(all of the ">/<" are underlined ones)

Answer 1

The range of the parabola y=-x^2+12x-11 with an axis of symmetry x=6 is y<25.

Step-by-step explanation:

The parabola y=-x^2+12x-11 has an axis of symmetry at x=6. The axis of symmetry is the vertical line that divides the parabola into two symmetric halves. The range of the parabola can be determined by finding the vertex. The vertex form of a parabola is given by y=a(x-h)^2+k, where (h,k) is the vertex. In this case, the vertex is (6, y) since the axis of symmetry is x=6. Plugging the x-coordinate 6 into the equation, we can find the y-coordinate:

y = -(6)^2 + 12(6) - 11 = -36 + 72 - 11 = 25

Therefore, the range of the parabola y=-x^2+12x-11 is y<25. So, the correct answer is option 2) y<25.