Calculate the coordinates of the points of intersection of the parabola y = 3x² + 6x and straight line = 6x.

Calculate the coordinates of the points of intersection of the parabola y = 3x² + 6x and straight line = 6x. in which coordinate quarters are these points? without drawing a schedule

1) We equate the right sides of the functions:

3x² + 6x = 6 – x.

2) Solve the quadratic equation:

3x² + 7x – 6 = 0.

D = 49 – (- 4 × 3 × 6) = 121.

X1 = (- 6 + √121) / 6 = 5/6 is the x-coordinate of the first intersection point.

X2 = (- 6 – √121) / 6 = – 17/6 = – 2 5/6 the x-coordinate of the second intersection point.

3) Find the coordinates along the y-axis:

y1 = 6 – 5/6 = 5 1/6.

y2 = 6 + 2 5/6 = 8 5/6.

Point A (5/6; 5 1/6) is in the first quarter, since both the abscissa and the ordinate are positive numbers.

Point B (- 2 5/6; 8 5/6) is in the second quarter, since the abscissa is negative and the ordinate is positive.