Question

write another polynomial function whose graph intercepts the horizontal axis at - 7, 8, and 15 and a different leading coefficient than your first function. how would the graph be different than your original? p(x)=2(x + 7)(x - 8)(x - 15)(x - 1)

Explanation:

Step1: Recall polynomial form

A polynomial with roots $r_1,r_2,r_3,r_4$ can be written as $y = a(x - r_1)(x - r_2)(x - r_3)(x - r_4)$. The given polynomial $P(x)=2(x + 7)(x - 8)(x - 15)(x - 1)$ has roots $x=-7,8,15,1$ and leading - coefficient $a = 2$.

Step2: Create new polynomial

To create a polynomial with the same roots but a different leading - coefficient, let's choose $a = 3$. The new polynomial is $Q(x)=3(x + 7)(x - 8)(x - 15)(x - 1)$.

Step3: Analyze graph differences

The leading - coefficient affects the end - behavior of the polynomial. If the leading coefficient $a>0$ and the degree of the polynomial is even (degree of $P(x)$ and $Q(x)$ is 4, which is even), as $x\to\pm\infty$, $y\to+\infty$. The magnitude of the leading coefficient affects the steepness of the graph. A larger leading coefficient makes the graph steeper. For $P(x)$ with $a = 2$ and $Q(x)$ with $a = 3$, the graph of $Q(x)$ will be steeper than the graph of $P(x)$ as $x$ moves away from the origin, but they will cross the $x$ - axis at the same points $x=-7,8,15,1$.

Answer:

A new polynomial is $Q(x)=3(x + 7)(x - 8)(x - 15)(x - 1)$. The graph of the new polynomial will be steeper than the graph of the original polynomial as $x$ moves away from the origin, but they have the same $x$ - intercepts.