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Chapter 3: Problem 86

Can the graph of a polynomial function have no \(x\) -intercepts? Explain.

Short Answer

Expert verified
Yes, a polynomial function can have no x-intercepts. This happens when all solutions of the equation P(x) = 0 are complex. An example of such a function is \(P(x) = x^2 + 1\), it has no real solutions, hence no x-intercepts.

Step by step solution

01

Understanding Polynomial Functions

A polynomial function can be expressed in the form \(P(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_2x^2 + a_1x + a_0\) where \(n\) is a nonnegative integer and \(a_0, a_1, \ldots, a_n\) are constants. \(a_n\) is called the leading coefficient and could not be zero.
02

Understanding Real Roots or X-intercepts

Roots or x-intercepts of the polynomial are the solutions of the equation \(P(x) = 0\). They correspond to the points where the graph of the polynomial function intersects the x-axis. A polynomial of degree \(n\) can have at most \(n\) real roots, or equivalently, \(n\) x-intercepts.
03

Polynomial Functions Without Real Roots or X-intercepts

For a polynomial function to have no real roots or x-intercepts, none of the solutions of the equation \(P(x) = 0\) can be real numbers. This happens when the polynomial is of even degree and all the solutions are complex, which means they are not real numbers. An example of such a function is \(P(x) = x^2 + 1\), which has no real solutions because the equation \(x^2 + 1 = 0\) implies that \(x^2 = -1\), which is not possible for any real number \(x\). Therefore, a solution can't be a real number, so the function has no x-intercepts.

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