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f(x)=x^2(x+2)(x-2)(x-5) has zeros at x=-2, x=0, x=2 and x=5. what is the sign of f on the interval -2

f(x)=x^2(x+2)(x-2)(x-5) has zeros at x=-2, x=0, x=2 and x=5. what is the sign of f-example-1
User Powkachu
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1 Answer

9 votes
9 votes

Answer:

f is sometimes positive and sometimes negative on the interval

Explanation:

So if you were to expand out the x's you would have x^2 * x * x * x which will result in x^5. and because the leading coefficient is positive, it will be going towards positive infinity as x goes towards positive infinity, and because the degree is odd the end behaviors will be opposite, so as x goes towards negative infinity the value of f(x) will be going towards negative infinity. One other thing to note is that because x^2 technically is two zeroes, with the same value, the zero at x=0 has a multiplicity of 2, meaning that it will not cross the x-axis, but rather have a turning point, once it intersects the x-axis, because the zero has an even multiplicity. Given this information we can draw a graph. So as you can see in the graph, it will sometimes be negative and sometimes positive

f(x)=x^2(x+2)(x-2)(x-5) has zeros at x=-2, x=0, x=2 and x=5. what is the sign of f-example-1
User Dmitry Volkov
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