Questions: Question 18 of 20 Graph the parabola. Give the vertex, axis of symmetry, domain, and range. f(x) = x^2 + 12x + 20 (Type an ordered pair.) The axis of symmetry is (Type an equation.) The domain is (Type your answer in interval notation.) The range is (Type your answer in interval notation.)

Find the vertex, axis of symmetry, domain, and range of the parabola \( f(x) = x^2 + 12x + 20 \).

Complete the square to find the vertex.

Rewrite \( f(x) = x^2 + 12x + 20 \) in vertex form by completing the square: \[ f(x) = (x^2 + 12x + 36) - 36 + 20 \] \[ f(x) = (x + 6)^2 - 16 \] The vertex form is \( f(x) = (x + 6)^2 - 16 \), so the vertex is \( (-6, -16) \).

Find the axis of symmetry.

The axis of symmetry is the vertical line that passes through the vertex. Therefore, the axis of symmetry is \( x = -6 \).

Determine the domain.

The domain of any quadratic function is all real numbers. Therefore, the domain is \( (-\infty, \infty) \).

Determine the range.

Since the parabola opens upwards (the coefficient of \( x^2 \) is positive), the range starts from the y-coordinate of the vertex and goes to infinity. Therefore, the range is \( [-16, \infty) \).

\(\boxed{\text{Vertex: } (-6, -16)}\) \(\boxed{\text{Axis of symmetry: } x = -6}\) \(\boxed{\text{Domain: } (-\infty, \infty)}\) \(\boxed{\text{Range: } [-16, \infty)}\)

\(\boxed{\text{Vertex: } (-6, -16)}\) \(\boxed{\text{Axis of symmetry: } x = -6}\) \(\boxed{\text{Domain: } (-\infty, \infty)}\) \(\boxed{\text{Range: } [-16, \infty)}\)

{"axisType": 3, "coordSystem": {"xmin": -10, "xmax": 0, "ymin": -20, "ymax": 10}, "commands": ["y = x**2 + 12*x + 20"], "latex_expressions": ["$f(x) = x^2 + 12x + 20$"]}