Question

MATH HELP PLEASE 50 POINTS:

Answer 1

Answer:

$f(x)=-a(x+2)(2x-1)(x-3)$ where a>0.

To graph the the polynomial, begin in the left top of quadrant 2. Then draw downwards to the first real zero on the x-axis at -2. Cross the x-axis and then curve back up to 1/2 on the x-axis. Cross through again and curve back down to cross for the last time at 3 on the x-axis. The graph then ends going down towards the right in quadrant 4. It forms an s shape.

Step-by-step explanation:

The real zeros are the result of setting each factor of the polynomial to zero. By reversing this process, we find:

• zero -2 is factor (x+2)

• zero 1/2 is factor (2x-1)

• zero 3 is factor (x-3)

We write them together with an unknown leading coefficient a which is negative so -a.

$f(x)=-a(x+2)(2x-1)(x-3)$ where a>0

The leading coefficient of a polynomial determines the direction of the graph's end behavior.

• A positive leading coefficient has the end behavior point up when an even degree and point opposite directions when an odd degree with the left down and the right up.

• A negative leading coefficient has the end behavior point down when an even degree and point opposite directions when an odd degree with the left up and the right down.

• This graph has all odd multiplicity. The graph will cross through the x-axis each time at its real zeros.

To graph the the polynomial, begin in the left top of quadrant 2. Then draw downwards to the first real zero on the x-axis at -2. Cross the x-axis and then curve back up to 1/2 on the x-axis. Cross through again and curve back down to cross for the last time at 3 on the x-axis. The graph then ends going down towards the right in quadrant 4. It forms an s shape.