a) y = x2 + 3x + 1b) y = 2x + 3At points of intersection (a) = (b).2x + 3 = x 2 + 3x + 1Note this is a quadratic expression which will solve for 2 unique solutions, providing the discriminant (b2 - 4ac, where a, b and c are the coefficients ax2 + bx + c = 0) is greater than 0, thus proving that there are two distinct points of intersection.Solve for x.x 2 + x - 2=0x2 - 2x + x - 2 = 0x ( x - 2 ) + 1 ( x - 2 ) = 0x + 1 = 0 , x - 2 = 0x = (-1) , x = 2Sub values into (b) to solve for y.y = 2x + 3y = 2(-1) + 3 , y = 2(2) + 3y = 1 , y = 7Use values for x and y to express points of intersection as co-ordinates in the form (x,y)Points of intersection are;(-1,1) and (2,7).