Question 1132242
<br>
(1) The curve touches the x-axis at x=1  -->  x=1 is a double root  -->  (x-1) is a factor twice<br>
(2) The curve passes through (-2,0)  -->  x = -2 is a root  -->  (x+2) is a factor<br>
So the polynomial is<br>
{{{f(x) = a(x-1)^2(x+2) = a(x^3-3x+2)}}}<br>
(3) The graph touches the line y=3x+6 at (-2,0)  -->  the slope (derivative of the function) evaluated at x=-2 is 3:<br>
{{{df/dx = a(3x^2-3)}}}
{{{a(3(-2)^2-3) = 3}}}
{{{a(12-3) = 3}}}
{{{9a = 3}}}
{{{a = 1/3}}}<br>
The polynomial is<br>
{{{f(x) = (1/3)(x-1)^2(x+2) = (1/3)(x^3-3x+2) = (1/3)(x-1)(x^2+x-2) = (x-1)((1/3)x^2+(1/3)x-(2/3))}}}<br>
ANSWER (a): In the form (x-1)(ax^2+bx+c), a = 1/3, b = 1/3, c = -2/3<br>
A graph (f(x) red; y=3x+6 green), showing the curve touching the x-axis at x=1 and touching the line y=3x+6 at (-2,0)...<br>
{{{graph(400,400,-4,4,-10,10,(1/3)(x-1)^2(x+2),3x+6)}}}<br>
ANSWER (b(2)): area under the curve from x=1 to x=4<br>
Taking the constant 1/3 outside the integral, we have<br>
1/3 of integral from 1 to 4 of ((x^3-3x+2)dx) is<br>
1/3 of {{{((1/4)(4^4)-(3/2)4^2+2(4))-((1/4)(1^4)-(3/2)1^2+2(1)) = ((64-24+8)-(1/4-3/2+2)) = (48-3/4) =189/4}}}<br>= 63/4