Question

write another polynomial function whose graph intercepts the horizontal axis at -7, 8, and 15 and a different degree than your first function. how would the graph be different than your original? p(x)=(x + 7)(x - 8)(x - 15)(x - 1)

Explanation:

Step1: Recall polynomial - root relationship

The roots of the polynomial \(p(x)=(x + 7)(x - 8)(x - 15)(x-1)\) are \(x=-7,8,15,1\) since when \(x\) takes these values, \(p(x)=0\). A polynomial function \(y = f(x)\) intersects the \(x\) - axis (horizontal axis) at its roots.

Step2: Create a new polynomial

A new polynomial with the same \(x\) - intercepts \(-7,8,15\) and an additional root. Let's consider a polynomial \(q(x)=(x + 7)(x - 8)(x - 15)(x - 0)\) (we added a root at \(x = 0\)). The degree of \(p(x)\) is \(4\) (since it is a product of 4 linear factors), and the degree of \(q(x)\) is \(5\) (a product of 5 linear factors).

Step3: Analyze the graph differences

The graph of \(q(x)\) will cross or touch the \(x\) - axis at \(x=-7,0,8,15\). Compared to \(p(x)\), the graph of \(q(x)\) has an additional \(x\) - intercept at \(x = 0\). Also, as the degree of \(q(x)\) is one more than the degree of \(p(x)\), the end - behavior of the two graphs is different. For an even - degree polynomial like \(p(x)\) (degree \(n = 4\)), as \(x\to\pm\infty\), \(y\to+\infty\). For an odd - degree polynomial like \(q(x)\) (degree \(n = 5\)), as \(x\to-\infty\), \(y\to-\infty\) and as \(x\to+\infty\), \(y\to+\infty\).

Answer:

A new polynomial could be \(q(x)=x(x + 7)(x - 8)(x - 15)\). The graph of \(q(x)\) has an additional \(x\) - intercept at \(x = 0\) compared to \(p(x)\), and the end - behavior is different due to the difference in degrees ( \(p(x)\) has degree 4 and \(q(x)\) has degree 5).