Linear equation: y = 3x+2 ……(1)

Quadratic equation: y = x2 ……(2)

(a)

Substitute x2 for y in the linear equation and solve

y=3x+2

\(x^2=3x+2\)

\(x^2-3x-2=0\)

\(x=(-(-3)pmsqrt((-3)^2-4(1)(-2))/(2(1))\)

\(x=3pmsqrt(17)/2\)

x=3.562 and x=0.562

substitute 3.562 for x in equation 2

\(y=x^2\)

\(y=(3.562)^2\)

y=12.685

substitute -0.562 for x in equation 2

\(y=x^2\)

\(y=(-0.562)^2\)

y=0.315

From above, the intersection points of line and curve are (-0.562, 0.315) and (3.562, 12.685).

(b)

The graph of line and curve is shown in the figure below.