Question

"Given \(y = -x^2 - 2x + 1\)

a. Axis of Symmetry:
a) \(x = -1\)
b) \(x = 1\)
c) \(x = -2\)
d) \(x = 2\)

b. Vertex:
a) \((-1, 0)\)
b) \((1, -2)\)
c) \((-1, 1)\)
d) \((-2, -1)\)

c. Y-intercept:
a) \((0, -1)\)
b) \((0, 1)\)
c) \((0, -2)\)
d) \((0, 2)\)

d. Domain:
a) \((-[infinity], [infinity])\)
b) \((-2, [infinity])\)
c) \((-[infinity], -2)\)
d) \((-1, [infinity])\)

e. Range:
a) \((-[infinity], 0]\)
b) \((-\infty, 1]\)
c) \((-[infinity], 2]\)
d) \((-1, [infinity])\)"

Answer 1

The axis of symmetry is x = 1, the vertex is (1, -2), the y-intercept is (0, 1), the domain is all real numbers, and the range is all real numbers less than or equal to 0.

Step-by-step explanation:

Let's analyze the quadratic function y = -x^2 - 2x + 1 and answer each part of the question:

a. Axis of Symmetry:

The axis of symmetry for a quadratic function in the form y = ax^2 + bx + c is given by x = -b/(2a). For our function, a = -1 and b = -2, so the axis of symmetry is x = -(-2)/(2(-1)) = -2/-2 = 1. Therefore, the correct option is b) x = 1.

b. Vertex:

To find the vertex, we use the axis of symmetry and plug it into the original equation. So, y = -(1)^2 - 2(1) + 1 = -1 - 2 + 1 = -2. Thus, the vertex is (1, -2), which corresponds to option b) (1, -2).

c. Y-intercept:

The y-intercept occurs when x = 0. Plugging x = 0 into the equation gives y = -(0)^2 - 2(0) + 1 = 1. So, the y-intercept is (0, 1), which is option b) (0, 1).

d. Domain:

The domain of any quadratic function is the set of all real numbers. Therefore, the correct domain is a) (-∞, ∞).

e. Range:

Since our quadratic function opens downward (a < 0), the maximum value is at the vertex. The range is all real numbers less than or equal to the y-coordinate of the vertex, which is -2. Thus, the range is a) (-∞, 0].