Graph this function, please?

Question

This is the formula:

y= x⁴ - 2x³ - 8x² +12x - 16

Please show all the steps? My exam is in a couple days and I am SO screwed...

Answer 1

Given function y = x^4-2x^3-8x^2+12x-16

Given 4th degree polynomial function , so the graph will behave roughly like a quadratic up on both endsor down on both ends.

All even degree polynomials behave,on thier ends, like quadratics.

Since the leading coefficent sign is positive the graph will be up on both ends.

Graph

From the graph x = 3.6, -2.8

Answer 2

The polynomial function is y = x⁴ - 2x³ - 8x² + 12x - 16.

To graph the polynomial function follow the steps :

• Find the zeros of a polynomial function.

The x - intercepts are the real zeros of the graph.

To find the x - intercepts substitute y = 0 in the given polynomial function.

0 = x⁴ - 2x³ - 8x² + 12x - 16.

To find x - intercepts, solve the above polynomial.

First find the roots of the polynomial.

Because, the degree  of the polynomial is 4, there are four roots to the polynomial.

The roots are .

The polynomial equation is .

Compare it to general quartic equation : .

.

.

Answer 3

Contd..........

.

So, the real zeroes(x - intercepts) are .

Answer 4

Contd.........

• find the test points of the polymonial.

Make the table of values for the polnomial function.

Here, i test 4 points to determine whether the graph of polynomial lies above or below the x axis.

Choose values for x and find the corresponding values for y.

x	y = x4 - 2x3 - 8x2 + 12x - 16	(x, y)
-1	y =(- 1)4 - 2(- 1)3 - 8(- 1)2 +12(- 1) - 16 = - 33	(- 1, - 33)
0	y =(0)4 - 2(0)3 - 8(0)2 +12(0) - 16 = - 16	(0, - 16)
1	y =(1)4 - 2(1)3 - 8(1)2 +12(1) - 16 = - 13	(1, - 13)
2	y =(2)4 - 2(2)3 - 8(2)2 +12(2) - 16 = - 24	(2, - 24)

• Use the leading coefficient test to find the end behavior of the polynomial function.

The polynomial function is y = x⁴ - 2x³ - 8x² + 12x - 16.

Degree of the polynomial is 4 and leading coefficient 1.

The graph of a polynomial function is always a smooth curve; that is, it has no breaks or corners.

All even degree polynomials behave on thier ends like squares.

All even degree polynomials  have ends that head off in same directions.depending on whether the polynomial has, respectively, a positive or negative leading coefficient.

The above polynomial is even degree  polynomial with a positive leading coefficient .

So the graph rises to the left and  to the right.

• Graph :

1.Draw a coordinate plane.

2.Plot the x - intercepts and coordinate points found in the table.

3.Then draw the graph, connecting the points with a smooth curve.