Identify the medium.y=x^2-4x-32
Medium is a nonstandard description of a part of the graph of a parabola.
Given `y=x^2-4x-32` the graph will be a parabola ( the function is a quadratic polynomial), opening up (it is a function and the leading coefficient is positive.)
The axis of symmetry can be found by `x=(-b)/(2a)` . In this case the axis of symmetry is `x=(-(-4))/(2(1))` or `x=2` .
The vertex lies on the axis of symmetry; if the x-coordinate is 2 then the y-coordinate is f(2) -- `f(2)=(2)^2-4(2)-32=-36` . The vertex is at (2,-36).
The x-intercepts can be found by factoring: `x^2-4x-32=(x-8)(x+4)==>(x-8)(x+4)=0 ==> x=-4,x=8`
Or you could use the quadratic formula:`x=(-b+-sqrt(b^2-4ac))/(2a)`
In this case `x=(4+-sqrt(16-4(-32)))/2==>x=2+-12/2` so x=8 or -4