Question 1116910
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If the curve touches the x-axis at x=1, then it has a double root there.  Then, since it intersects the line y=3x+6 at (-2,0), the equation for the function is<br>
{{{a(x-1)^2(x+2)}}}<br>
where a can be any nonzero constant.<br>
The rest of the problem as shown can't be completed, because there was no information to use to evaluate the constant.<br>
So to continue, I will assume the constant is 1; then the equation of the function is<br>
{{{(x-1)^2(x+2)}}}<br>
Since the description of the problem said the equation was {{{(x-1)(ax^2+bx+c)}}}, it must be that {{{ax^2+bx+c = (x-1)(x+2) = x^2+x-2}}}.  So the answers for part 1 of the problem are
a=1; b=1; c=-2<br>
Here is the graph for part 2:<br>
{{{graph(400,400,-3,3,-10,20,(x-1)^2*(x+2),3x+6)}}}<br>
The x values of the points of intersection are 0, 1-sqrt(3), and 1+sqrt(3).  You can find the last two by solving the pair of equations simultaneously, knowing that one of the intersection points is at (-2,0).<br>
I will let you do the ugly integration to find the area of the finite regions between the two curves.