Which factor could be part of the function so that the function is increasing on the left and decreasing on the right? Select allthat apply.f(x) = (x+4)(- 5)Options are:-0.8-3x(x+5)-(2x+1)(3x^2+5)

Answer:Only CStep-by-step explanation:Recall from the properties of end behaviors of polynomials, that:a) if the lead coefficient is negative, the end behavior on the right side has a negative slope (and vice versa), also....b) if the degree is even, then the end behavior of the left side must be pointing in the same vertical direction as the right side.Back to the question:Hence to have the left side increasing and the right side decreasing, we can conclude that based on the above properties, that.a) since the right side is decreasing (negative slope), hence the polynomial must have a lead coefficient that is negative.b) since one side of the graph is increasing, and the other side is decreasing, the function must look roughly like an "n" shape, which means that the end behavior of both the left and right side points downward. Hence the degree of the polynomial is EVEN.so....f(x) = (x+4)(- 5)f(x) = -5x - 20all we need to do is go down the list and multiply f(x) with the options and see which one fits the 2 criteria we determined above.A) f(x) . (-0.8) = (-5x -20)(-0.8) = 4x + 16 (positive lead coeff. and odd degree, Does Not Aply)B) f(x) . (-3x) = (-5x -20)(-3x) = 15x² + 60 (positive lead coeff. and even degree, Does not apply)C) f(x) . (x+5) = (-5x -20)(x+5) = -5x² - 45x - 100 (negative lead coeff, even degree THIS APPLIES!)I'm not going to work out the last 2, but they giveD) f(x) . -(2x+1) = 10x²+45x+20 (positive lead coeff, even degree, Does not apply)E) f(x) . (3x²+5) = −15x³−60x²−25x−100 (negative lead coeff, odd degree, Does not apply)Hence only C applies