Question

1. using the axis to the right, draw an example of a polynomial function that matches the statements $lim_{x

ightarrow-infty}p(x)=infty$ and $lim_{x
ightarrowinfty}p(x)=-infty$.

Explanation:

Step1: Analyze end - behavior

We know that for a polynomial \(y = p(x)=a_nx^n + a_{n - 1}x^{n-1}+\cdots+a_1x + a_0\), if \(n\) is odd and \(a_n>0\), \(\lim_{x
ightarrow\infty}p(x)=\infty\) and \(\lim_{x
ightarrow-\infty}p(x)=-\infty\). If \(n\) is odd and \(a_n < 0\), \(\lim_{x
ightarrow\infty}p(x)=-\infty\) and \(\lim_{x
ightarrow-\infty}p(x)=\infty\). The given conditions \(\lim_{x
ightarrow\infty}p(x)=\infty\) and \(\lim_{x
ightarrow-\infty}p(x)=-\infty\) imply that the polynomial has an odd degree with a positive leading coefficient.

Step2: Sketch the polynomial

Start from the bottom - left quadrant (since \(\lim_{x
ightarrow-\infty}p(x)=-\infty\)). Draw a smooth curve that crosses the \(x\) - axis zero, one, two or more times (depending on the number of real roots of the polynomial) and then goes up towards the top - right quadrant (since \(\lim_{x
ightarrow\infty}p(x)=\infty\)). A simple example of such a polynomial is \(y=x^3 - x\). To draw it:

1. Find the \(x\) - intercepts by setting \(y = 0\), so \(x^3 - x=x(x - 1)(x + 1)=0\), and the \(x\) - intercepts are \(x=-1,0,1\).
2. The \(y\) - intercept is found by setting \(x = 0\), so \(y = 0\).
3. Sketch a smooth curve passing through these points, starting from the bottom - left and ending at the top - right.

Answer:

Draw a smooth curve starting from the bottom - left quadrant, crossing the \(x\) - axis at some points (e.g., \(x=-1,0,1\) for \(y=x^3 - x\)) and going to the top - right quadrant.