Question

problems 15 - 18, use the factored form of ( g(x) ) to identify key characteristics and sketch.
( g(x)=-\frac{1}{2} x^{2}(x + 1)(x^{2}-4) )

1. zeros of and their multiplicities
2. ( y )-intercept
3. end behavior - use limit notation
4. sketch the graph of ( g(x) ).

Explanation:

Step1: Find the zeros

Set \(g(x) = 0\).
\(-\frac{1}{2}x^{2}(x + 1)(x^{2}-4)=0\).
Since \(x^{2}-4=(x - 2)(x + 2)\), the equation becomes \(-\frac{1}{2}x^{2}(x + 1)(x - 2)(x + 2)=0\).
By the zero - product property, \(x = 0\) (multiplicity \(2\)), \(x=-1\) (multiplicity \(1\)), \(x = 2\) (multiplicity \(1\)), \(x=-2\) (multiplicity \(1\)).

Step2: Find the \(y\) - intercept

Set \(x = 0\) in \(g(x)\).
\(g(0)=-\frac{1}{2}(0)^{2}(0 + 1)(0^{2}-4)=0\).

Step3: Determine the end - behavior

The leading term of \(g(x)\) is found by multiplying the leading terms of each factor.
The degree of \(g(x)\) is \(2+1 + 2=5\) (odd) and the leading coefficient \(a=-\frac{1}{2}\).
\(\lim_{x
ightarrow\infty}g(x)=-\infty\) (because \(a<0\) and \(n = 5\) (odd)) and \(\lim_{x
ightarrow-\infty}g(x)=\infty\) (because \(a<0\) and \(n = 5\) (odd)).

Answer:

1. Zeros: \(x = 0\) (multiplicity \(2\)), \(x=-1\) (multiplicity \(1\)), \(x = 2\) (multiplicity \(1\)), \(x=-2\) (multiplicity \(1\)).
2. \(y\) - intercept: \((0,0)\).
3. End - behavior: \(\lim_{x

ightarrow\infty}g(x)=-\infty\), \(\lim_{x
ightarrow-\infty}g(x)=\infty\).

1. To sketch the graph:

• Mark the \(x\) - intercepts at \(x=-2,-1,0,2\).
• Since \(x = 0\) has multiplicity \(2\), the graph touches the \(x\) - axis at \(x = 0\).
• For \(x=-2,-1,2\) (multiplicity \(1\)), the graph crosses the \(x\) - axis.
• Use the end - behavior \(\lim_{x

ightarrow\infty}g(x)=-\infty\) and \(\lim_{x
ightarrow-\infty}g(x)=\infty\) to draw the ends of the graph.