Respuesta :

Answer:

option-B

Step-by-step explanation:

We are given function as

[tex]f(x)=-x^4+3x^3+10x^2[/tex]

we know that

Any function touches or crosses x-axis when y-value is 0

so, we can set f(x)=0

and then we can solve for x

[tex]f(x)=-x^4+3x^3+10x^2=0[/tex]

now, we can factor it

[tex]-x^2(x^2-3x-10)=0[/tex]

[tex]-x^2(x-5)(x+2)=0[/tex]

we get

[tex]-x^2=0[/tex]

[tex]x=0[/tex]

It means that function touches x-axis at x=0

[tex](x-5)(x+2)=0[/tex]

[tex](x-5)=0[/tex]

[tex]x=5[/tex]

[tex](x+2)=0[/tex]

[tex]x=-2[/tex]

So, function crosses x-axis at x=5 and x=-2

so,

option-B

Answer:

The graph touches the x-axis at x=0 and crosses the x-axis at x=5 and x=-2.

Step-by-step explanation:

We have the following function and we are to solve it for x:

[tex]f(x)=-x^4+3x^3+10x^2[/tex]

We will put the given function equal to zero and solve it as we know that the function crosses the x-axis when y = 0.

[tex]-x^4+3x^3+10x^2=0[/tex]

So taking the common terms out:

[tex]-x^2(x^2-3x-10)=0[/tex]

[tex]-x^{2} (x-5)(x+2)=0[/tex]

[tex]-x^{2} =0, x=0[/tex]

[tex](x-5)=0, x=5[/tex]

[tex](x+2)=0, x=-2[/tex]

Therefore, the graph touches the x-axis at x=0 and crosses the x-axis at x=5 and x=-2.

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