# how to determine the concavity, x-intercpt and y intercept

f(x)=-3(x+5)(x-11)

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urvashiparmar
30 August 2017
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urvashiparmar
30 August 2017
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urvashiparmar
30 August 2017
The answer is provided in the images
urvashiparmar
30 August 2017
for concavity check function second order derivative  if it is positive then it is concave up othrerwise it is concave down.dy/dx=18-6xd2y/d2x=-6so it is negative means it is concave down functon.x-intercept (-5,0) and (11,0)y-intercept (0,165)
asnitkkr21
30 August 2017
As far as the concavity is concerned you can obviously differentiate (as a colleague already did), or you can instantly observe that the quadratic equation that you have is -3x^2+...... (I don't care about the rest). The coefficient of x^2 is a negative number, this instantly means that we have an "n" curve, therefore a concave down function. As far as the x-intercept is concerned, we already know that on the x-axis, y is always zero, and y=f(x). -3(x+5)(x-11)=0.  We have a product that involves two brackets. Since -3 cannot be zero, inevitably one of the brackets has to be equal to zero, which means x+5=0 or x-11=0 which gives you x=-5 or x=11. Thus, you have the points A(-5,0) and B(11,0)Similar thing with the y- intercept. On the y-axis all the x's are zero, so you just substitute x by zero which gives you f(0)=-3(0+5)(0-11)=165 which means your function meets the y-axis at point C(0,165)
marias
08 September 2017