# f(x) = x^2+11x-5

## The vertex is located at -5.5, -35.25

I found the vertex using the vertex formula, a(x-h)^2+k, where x and y equal h and k respectively.
(11/2)^2 = (5.5)^2 = 30.25
x^2+11x+30.25-30.25-5
(x+5.5)^2-30.25-5
(x+5.5)^2-35.25

## The Axis of Symmetry is -5.5

I found the axis of symmetry using the vertex formula, a(x-h)^2+k, where h equals the x intercept.
(11/2)^2 = (5.5)^2 = 30.25
x^2+11x+30.25-30.25-5
(x+5.5)^2-30.25-5
(x+5.5)^2-35.25

## The minimum value is -35.25

I found the minimum value using the vertex formula, a(x-h)^2+k, where k equals the minimum value.
(11/2)^2 = (5.5)^2 = 30.25
x^2+11x+30.25-30.25-5
(x+5.5)^2-30.25-5
(x+5.5)^2-35.25

## The y-intercept is -5

I found the y-intercept by substituting 0 for x in the vertex form of the quadratic function.
(11/2)^2 = (5.5)^2 = 30.25
x^2+11x+30.25-30.25-5
(x+5.5)^2-30.25-5
(x+5.5)^2-35.25
(0+5.5)^2-35.25
5.5^2-35.25
30.25-35.25
-5

## The x-intercepts are .44 and -11.44

I solved the equation by using the quadratic formula to find the x-intercepts.
x = -11 +-sqrt11^2-4(1)(-5) / 2(1)
x = -11 +-sqrt121-(-20) / 2
x = -11 +-sqrt141 / 2
x = -11 +-11.87 / 2
x = -5.5 +-5.935
x = 0.435 or x = -11.435
Additionally, because there are no like terms, this equation cannot be factored.

## Other Points

I know immediately after finding the y-intercept that due to the reflective property of symmetry, the parabola intersects a point that is the same distance from the axis of symmetry. This point is -11, -5. Additionally, from looking at the graph, I can see that the parabola intersects point -10, -15 and another point from reflection across the axis of symmetry at -1,-15.