Answer the following quadratic function

f(x)= x^2 + 6x - 16
a)What is the vertex (h,k)of f?
b)What is the axis of symmetry?
c)What are the intercepts?

x^2 + 6x - 16 = (x+3)^2 - 25 = (x+8)(x-2)

that should get you started.

2,-8 is your answer

To find the vertex, axis of symmetry, and intercepts of the quadratic function f(x) = x^2 + 6x - 16:

a) The vertex of a quadratic function in the form f(x) = ax^2 + bx + c is given by the coordinates (h, k), where h = -b/2a and k = f(h).

In this case, a = 1, b = 6, and c = -16.
Substituting these values into the formula, we get:
h = -6 / (2 * 1) = -6 / 2 = -3
f(-3) = (-3)^2 + 6(-3) - 16 = 9 - 18 - 16 = -25

Therefore, the vertex of the function f(x) is (-3, -25).

b) The axis of symmetry is a vertical line that passes through the vertex. Its equation is given by x = h, where h is the x-coordinate of the vertex.

In this case, the axis of symmetry is x = -3.

c) To find the x-intercepts, we need to set f(x) = 0 and solve for x.

Setting f(x) = x^2 + 6x - 16 = 0, we can use factoring, completing the square, or the quadratic formula to solve for x.

Using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
Substituting the values a = 1, b = 6, and c = -16 into the formula, we get:
x = (-6 ± √(6^2 - 4 * 1 * (-16))) / (2 * 1)
x = (-6 ± √(36 + 64)) / 2
x = (-6 ± √(100)) / 2
x = (-6 ± 10) / 2

Therefore, the x-intercepts are:
x = (-6 + 10) / 2 = 4 / 2 = 2
x = (-6 - 10) / 2 = -16 / 2 = -8

Therefore, the x-intercepts of the function f(x) are x = 2 and x = -8.

To find the y-intercept, we need to evaluate f(x) when x = 0.
f(0) = (0)^2 + 6(0) - 16 = -16

Therefore, the y-intercept of the function f(x) is -16.

In summary:
a) The vertex of f is (-3, -25).
b) The axis of symmetry is x = -3.
c) The x-intercepts are x = 2 and x = -8, and the y-intercept is (0, -16).

To answer these questions, we'll use the general form of a quadratic function: f(x) = ax^2 + bx + c.

a) To find the vertex (h, k) of the quadratic function, we can use the formula h = -b/2a and then substitute the value of h into the equation to find k.

In this case, a = 1, b = 6, and c = -16. Substituting these values into the formula, we have:
h = -6 / (2 * 1) = -6 / 2 = -3

Now substitute h = -3 into the equation:
f(-3) = (-3)^2 + 6(-3) - 16
f(-3) = 9 - 18 - 16
f(-3) = -25

Therefore, the vertex of function f is (-3, -25).

b) The axis of symmetry is the vertical line that passes through the vertex of the parabola. It can be found using the x-coordinate of the vertex.

In this case, the x-coordinate of the vertex is -3, so the equation of the axis of symmetry is x = -3.

c) To find the x-intercepts (zeros) of the quadratic function, we set f(x) = 0 and solve for x. The x-intercepts occur when the function crosses or touches the x-axis.

Setting f(x) = 0, we have:
x^2 + 6x - 16 = 0

To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula, which states that if ax^2 + bx + c = 0, then x = (-b ± √(b^2 - 4ac)) / (2a).

Using our values from the quadratic equation, a = 1, b = 6, and c = -16, we have:
x = (-6 ± √(6^2 - 4 * 1 * -16)) / (2 * 1)
x = (-6 ± √(36 + 64)) / 2
x = (-6 ± √(100)) / 2
x = (-6 ± 10) / 2

This gives us two possible solutions:
x1 = (-6 + 10) / 2 = 4 / 2 = 2
x2 = (-6 - 10) / 2 = -16 / 2 = -8

Therefore, the x-intercepts of function f are x = 2 and x = -8.

To find the y-intercept, we substitute x = 0 into the equation:
f(0) = 0^2 + 6(0) - 16
f(0) = -16

So, the y-intercept is y = -16.