Without constructing, find the coordinates of the intersection points of the parabola y = x2-14
Without constructing, find the coordinates of the intersection points of the parabola y = x2-14 and the straight line x + y = 6
The intersection points of the graphs of functions belong to both one graph and another graph. Therefore, in order to find their coordinates, it is necessary to combine the equations into a system and solve it.
{y = x ^ 2 – 14; x + y = 6 – express from the second equation y through x;
{y = x ^ 2 – 14; y = 6 – x – because the left sides of the equations are equal, then we equate their right sides;
x ^ 2 – 14 = 6 – x;
x ^ 2 + x – 14 – 6 = 0;
x ^ 2 + x – 20 = 0;
D = b ^ 2 – 4ac;
D = 1 ^ 2 – 4 * 1 * (-20) = 1 + 80 = 81; √D = 9;
x = (-b ± √D) / (2a);
x1 = (-1 + 9) / 2 = 8/2 = 4;
x2 = (-1 – 9) / 2 = -10/2 = -5;
y1 = 6 – x1 = 6 – 4 = 2;
y2 = 6 – x2 = 6 – (-5) = 6 + 5 = 11.
Answer. (4; 2); (-5; 11).