Let f be the function defined by f(x)=ax^2+bx+2\2x^2−8, where a and b are constants. the graph of f has a horizontal asymptote at y=3, and f has a removable discontinuity at x=2.

(c) at what value of x does f have a discontinuity due to a vertical asymptote? give a reason for your answer.

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step-by-step explanation:

honesty mood

step-by-step explanation:

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