3.7k views
1 vote
Identify the vertex, axis of symmetry, domain, range, and x- and y-intercepts for the following quadratic functions.

Function: f(x) = −2(x + 1)^2 + 8

1 Answer

0 votes

Final answer:

The vertex of the quadratic function f(x) = −2(x + 1)2 + 8 is (-1, 8), the axis of symmetry is x = -1, the domain is all real numbers, the range is y ≤ 8, there are no x-intercepts, and the y-intercept is (0, 6).

Step-by-step explanation:

The student asked to identify the vertex, axis of symmetry, domain, range, and x- and y-intercepts of the function f(x) = −2(x + 1)2 + 8. This function is a quadratic equation in vertex form where the standard form is f(x) = a(x - h)2 + k, where (h,k) is the vertex of the parabola.

  • The vertex can be identified by looking at the values of h and k in the function's vertex form. Here, the vertex is (-1, 8).
  • The axis of symmetry of a parabola is the vertical line that passes through its vertex, which can be described by the equation x = h. Therefore, for this function, the axis of symmetry is x = -1.
  • The domain of any quadratic function is all real numbers because there are no restrictions on the values x can take.
  • The range of the function, since the leading coefficient a = -2 is negative, is all real numbers less than or equal to the y-value of the vertex, which is y ≤ 8.
  • To find the x-intercepts, set f(x) to zero and solve for x. Since the function is in vertex form and the vertex has a y-value greater than zero, it is evident that there are no real x-intercepts for this function.
  • The y-intercept is found by setting x to zero. For the function f(x) = -2(x + 1)2 + 8, by substituting x with 0, we find the y-intercept to be (0, 6).
User Fobus
by
7.0k points