<PRE><B>Question 1128653</B>
&lt;br&gt;
y = ax^3+bx^2+cx+d
y&#39; = 3ax^2+2bx+c
y&#39;&#39; = 6ax+2b&lt;br&gt;
The given information tells us...&lt;br&gt;
y(0) = 15/2  --&gt;  c = 15/2  (1)
y(3) = 6  --&gt;  27a+9b+3c+d = 6 (2)
y&#39;(0) = -3  --&gt;  c = -3  (3)
y&#39;&#39;(3) = 0  --&gt;  18a+2b = 0  (4)&lt;br&gt;
Substitute (1) and (3) in (2):&lt;br&gt;
27a+9b-9+15/2 = 6
27a+9b = 15/2
18a+6b = 5 (5)&lt;br&gt;
Subtract (4) from (5):&lt;br&gt;
4b = 5
b = 5/4  (6)&lt;br&gt;
Substitute (6) in (4):&lt;br&gt;
18a+5/2 = 0
a = -5/36&lt;br&gt;
The polynomial is&lt;br&gt;
{{{(-5/36)x^3+(5/4)x^2-3x+(15/2)}}}&lt;br&gt;
A graph...&lt;br&gt;
{{{graph(400,400,-2,10,-2,10,(-5/36)x^3+(5/4)x^2-3x+(15/2))}}}</PRE>