Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,459 questions

17,854 answers

1,446 comments

807,720 users

graphically find the zeros of y=6x^4+55x^3+50x^2-275x+84

0 votes
sketch a graph of the answers.
asked Mar 10, 2014 in ALGEBRA 2 by payton Apprentice

1 Answer

0 votes

The polynomial y = 6x 4 + 55x 3 + 50x 2 - 275x + 84

Real zeros are intercepts of the graph.

1)Test points

Make the table of values to for the polnomial.

Here i test 5 points to determine whether the graph of polynomials lies above or below the x axis.

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

x

y = 6x 4 + 55x 3 + 50x 2 - 275x + 84 (x, y )

- 6

y = 6(-6)4 + 55(-6) 3 + 50(-6) 2 - 275(-6)+ 84

    y = -570

(- 6, -570)

-4

y = 6(-4)4 + 55(-4) 3 + 50(-4) 2 - 275(-4)+ 84

 y  = 0

(-4, 0)

0

y = 6(0)4 + 55(0) 3 + 50(0) 2 - 275(0)+ 84

y  = 84

(0, 84)
2

y = 6(2)4 + 55(2) 3 + 50(2) 2 - 275(2)+ 84

    y  = 270

(2, 270)
1

y = 6(1)4 + 55(1) 3 + 50(1) 2 - 275(1)+ 84

 y = -80

(1, -80)

2) End behavior y = 6x 4 + 55x 3 + 50x 2 - 275x + 84

Degree of polynomial is 4 and leading coefficient 6.

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 quadratics.

All even degree polynomials are either up on both ends and or down on both ends.depending on whether the polynomial has, respectively, a positive or negative leading coefficient.

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

So the graph up on both ends.

3)Graph

1.Draw a coordinate plane.

2.Plot the coordinate points found in the table.

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

From the graph we can observe the zeros of  y = 6x 4 + 55x 3 + 50x 2 - 275x + 84 are

x = - 7 , - 4 , 0.3 and 1.5.

answered Apr 7, 2014 by david Expert

Related questions

...