A quadratic equation of the form ax^2+bx+c = 0 has two different real roots. This is possible if the corresponding parabola has:
a) a > 0 and a vertex of (3,-2)
b)a > 0 and a vertex of (-3,2)
c) a < and a vertex of (-3,-2)
d) a < and a vertex of (3,0)

answer: -8.2562234e+13

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204