SOLUTION: 1. Let A(x)=x^3-5x^2-8x+40 and B(x)=3x^2-10x-8 . a) Determine the zeroes of B(x) algebraically. b) Draw the graph of A(x) . What occurs at the x-values found in (a)? c) Ho

Algebra ->  Rational-functions -> SOLUTION: 1. Let A(x)=x^3-5x^2-8x+40 and B(x)=3x^2-10x-8 . a) Determine the zeroes of B(x) algebraically. b) Draw the graph of A(x) . What occurs at the x-values found in (a)? c) Ho      Log On


   

Question 1037240: 1. Let A(x)=x^3-5x^2-8x+40 and B(x)=3x^2-10x-8 .
a) Determine the zeroes of B(x) algebraically.
b) Draw the graph of A(x) . What occurs at the x-values found in (a)?
c) How are the coefficients and exponents in B(x) related to the coefficients and exponents in A(x) ?
d) Use your observations in (c) to find the b(x) function if A(x)= 2x^3-7x^2+4x , and then use your observations in (b) to describe the shape of the graph of A(x) by finding the zeroes of B(x) .
Answer by josgarithmetic(39838) About Me  (Show Source):
You can put this solution on YOUR website!
A is a cubic, degree-three function, and B is a quadratic function.

3x%5E2-10x-8=0
%283x%2B2%29%28x-4%29=0
system%28x=-2%2F3%2Cor%2Cx=4%29

How does the derivative look for A?
x%5E3-5x%5E2-8x%2B40
dA%2Fdx=3x%5E2-10x-8, and you already know the zeros for this because it is the same function expression as B(x).

The slope of A is 0, both at x=-2/3, and at x=4.
That means A has extreme points (highest or lowest values) at those x values.
Which type of extreme, you can check through second derivative and testing points near the extremes.

Just to cut out some of the work, Here is A(x).
graph%28300%2C300%2C-4%2C6%2C-8%2C45%2Cx%5E3-5x%5E2-8x%2B40%29