Question

Y=x^2 - 2x+3 y=0;x=0;x=2

Answer 1

Ответ: x=1+i√2,1−i√2; у=3; у=3.

Объяснение:

y=х² - 2x+3;

y=0;

0=х²-2х+3;

x=1+i√2,1−i√2.

х=0;

y=х² - 2x+3;

у=0²-2×0+3;

у=3.

х=2;

у=х² - 2x+3;

у=2²-2×2+3;

у=3.

Answer 2

It seems like you want to find the points of intersection between the quadratic equation y = x^2 - 2x + 3 and the lines y = 0, x = 0, and x = 2. Let's solve for these points of intersection one by one:

1. Quadratic Equation and y = 0 (x-axis):

   To find the points where the quadratic curve crosses the x-axis (y = 0), you need to set y to 0 and solve for x:

   0 = x^2 - 2x + 3

   This is a quadratic equation. You can solve it using the quadratic formula:

   x = (-b ± √(b^2 - 4ac)) / (2a)

   In this case, a = 1, b = -2, and c = 3. Plugging these values into the formula:

   x = (-(-2) ± √((-2)^2 - 4(1)(3))) / (2(1))

   x = (2 ± √(4 - 12)) / 2

   x = (2 ± √(-8)) / 2

   Since the discriminant (the value inside the square root) is negative, there are no real solutions for x in this case. This means the quadratic curve does not intersect the x-axis.

2. Quadratic Equation and x = 0 (y-axis):

   When x = 0, the equation y = x^2 - 2x + 3 becomes:

   y = 0^2 - 2(0) + 3 y = 0 + 0 + 3 y = 3

   So, when x = 0, y = 3. This gives you a point at (0, 3) on the graph.

3. Quadratic Equation and x = 2:

   When x = 2, the equation y = x^2 - 2x + 3 becomes:

   y = 2^2 - 2(2) + 3 y = 4 - 4 + 3 y = 0 + 3 y = 3

   So, when x = 2, y = 3. This gives you a point at (2, 3) on the graph.

In summary, the quadratic curve y = x^2 - 2x + 3 does not intersect the x-axis (y = 0), but it intersects the y-axis (x = 0) at the point (0, 3) and the vertical line x = 2 at the point (2, 3).