f(x)=4x^2(x-2)^3 x-intercepts,y-intercepts end behavior

Question

function  F(x)= 4x^2 (x-2)^3  x-intercepts, y-intercepts and end behavior of the graph.

Answer 1

The polynomial function f (x ) =  4x ² ( x - 2)³

1) Zeros f (x ) =  4x ² ( x - 2)³ using zeros to graph the polynomial.

y  =  4x ² ( x - 2)³

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

4x ² ( x - 2)³   = 0

Apply zero product property.

4x ² = 0 and ( x - 2)³ = 0

x ² = 0 and  x - 2 = 0

x  = 0 and x  = 2

Reall zeros of f (x ) =  4x ² ( x - 2)³ are 0 , 2.

Real zeros are x  intercepts of the graph.

To find y  intercept substitute x  = 0 in the  function.

y  =  4(0)² ( 0 - 2)³

y = 0

2)Test points

Make the table of values to for the polnomial.

Here i test 3 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  =  4x 2 ( x - 2)3	(x, y )
- 0.5	y  =  4( -0.5)2(-0.5 - 2)3= -15.625	(- 0.5, -15.625)
0	y  =  4( 0)2(0 - 2)3= 0	(0, 0)
0.5	y = 4( 0.5)2(0.5 - 2)3= -3.375	(0.5, -3.375)

3) End behavior f (x ) =  4x ² ( x - 2)³

f (x ) = 4x ²( x ³ - 6x ² + 12x - 8)

f (x ) = 4x ⁵- 24x ⁴ + 48x ³ - 32x ²

Degree of the polynomial is 5 and leading coefficient 4.

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

All odd degree polynomials behave on thier ends like cubics.

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

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

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

4)Graph

1.Draw a coordinate plane.

2.Plot the x  intercepts 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 x - intercepts of the function (0,0) and (0,2).

y - intercept is (0, 0).