# f(x)=  x4 + x3 – 13x2 – 1x + 12  f(x)=x^4 + x^3 - 13x^2 - 1x+12 graph the following polynomial graphs on graph paper.   You will need to include:  end behavior, x-intercepts with...

f(x)=  x4 + x3 – 13x2 – 1x + 12

f(x)=x^4 + x^3 - 13x^2 - 1x+12

graph the following polynomial graphs on graph paper.

You will need to include:  end behavior, x-intercepts with the touch and turn around OR cross the x-axis, and y-intercept.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll determine x and y intercepts.

The graph is intercepting x axis for y = 0.

Since y = f(x) => x^4 + x^3 - 13x^2 - x + 12 = 0

Since all the coefficients of the given function are integer numbers, the roots of the function can be found among the divisors of the free coefficient, 12.

We'll check if x = -4 is one of the roots.

f(-4) = 256 - 64 - 208 + 4 + 12

f(-4) = 0 => x = -4 is one of the roots of the function.

Applying long division, we'll get the quotient q = x^3 - 3x^2 - x + 3.

We'll apply the reminder theorem:

(x + 4)(x^3 - 3x^2 - x + 3) = f(x)

If f(x) = 0 => (x + 4)(x^3 - 3x^2 - x + 3)

(x + 4)[x^2*(x - 3) - (x - 3)] = 0

(x + 4)(x - 3)(x^2 - 1) = 0

The difference of squares returns the product:

x^2 - 1 = (x - 1)(x + 1)

(x + 4)(x - 3)(x - 1)(x + 1) = 0

We'll cancel each factor:

x + 4 = 0 => x = -4

x - 3 = 0 => x = 3

x - 1 = 0 => x = 1

x + 1 = 0 => x = -1

The graph is intercepting x axis in the points: (4, 0), (3, 0), (-1, 0), (1, 0).

The graph  is intercepting y axis if x = 0.

f(0) = 0^4 + 0^3 – 13*0^2 – 0 + 12

f(0) = 12

The graph  is intercepting y axis in the point (0, 12).

The end behavior of the function is found when is evaluated the limit of the function if x is approaching to +/-infinite:

x approaches to + infinite:

lim x^4 + x^3 - 13x^2 - x+12 = (+infinite)^4 = +infinite

x approaches to - infinite:

lim x^4 + x^3 - 13x^2 - x+12 = (-infinite)^4 = +infinite