Quadratic Equation: \(y=-x^2+12x-11\)
What is the Vertex?
Vertex Form: \(y=a(x - h)^2 + k\), where (h, k) is the vertex of the parabola.
\(y=-x^2+12x-11\)
\(y=(-x^2+12x)-11\)
\(y=-1(x^2-12x)-11\)
\(y=-1(x^2-12x+36-36)-11\)
\(y=-1(x^2-12x+36)+36-11\)
\(y=-1(x-6)^2+25\)
Vertex is (6,25)
Axis of Symmetry?
This is just the h value so the AoS = 6
Total Distance? (I assume x-intercepts)
\(y=-x^2+12x-11\)
\(y=-1(x^2-12x+11)\)
\(y=-(x-1)(x-11)\)
x-intercepts are (1,0) and (11,0) so in between them was the flight.
Domain and Range?
I'm not sure what domain and range are but I found a calc online for it and it says
Domain: (−∞,∞),{x|x∈R}Range: (−∞,25],{y|y≤25}
