Question

Найдите p и постройте график функции y=x^2+p, если известно, что прямая y=-x имеет с графиком ровно одну общую точку.

Answer 1

Переносим х² и получаем p=y-x²

Answer 2

Problem Analysis

We are given the equation of a line, y = -x, and we need to find the value of p and graph the function y = x^2 + p such that the line and the graph have exactly one common point.

Solution

To find the value of p, we need to find the x-coordinate of the common point between the line y = -x and the graph y = x^2 + p.

Setting the two equations equal to each other, we have:

-x = x^2 + p

Simplifying the equation, we get:

x^2 + x + p = 0

This is a quadratic equation in terms of x. For the quadratic equation to have exactly one solution, the discriminant (b^2 - 4ac) must be equal to zero.

In this case, a = 1, b = 1, and c = p. So, the discriminant is:

(1)^2 - 4(1)(p) = 1 - 4p

Setting the discriminant equal to zero, we have:

1 - 4p = 0

Solving for p, we get:

4p = 1

p = 1/4

Therefore, the value of p is 1/4.

Now, let's graph the function y = x^2 + 1/4.

Graph of y = x^2 + 1/4

To graph the function y = x^2 + 1/4, we can start by plotting a few points and then connecting them to form a smooth curve.

Let's choose some x-values and calculate the corresponding y-values:

When x = -2, y = (-2)^2 + 1/4 = 4 + 1/4 = 17/4

When x = -1, y = (-1)^2 + 1/4 = 1 + 1/4 = 5/4

When x = 0, y = (0)^2 + 1/4 = 0 + 1/4 = 1/4

When x = 1, y = (1)^2 + 1/4 = 1 + 1/4 = 5/4

When x = 2, y = (2)^2 + 1/4 = 4 + 1/4 = 17/4

Now, let's plot these points on a graph:

| x | y | |---|---| | -2 | 17/4 | | -1 | 5/4 | | 0 | 1/4 | | 1 | 5/4 | | 2 | 17/4 |

We can see that the graph is symmetric with respect to the y-axis and opens upward. The vertex of the parabola is at (0, 1/4), and the axis of symmetry is the y-axis.

Conclusion

The value of p is 1/4, and the graph of the function y = x^2 + 1/4 has exactly one common point with the line y = -x. The graph is symmetric with respect to the y-axis, opens upward, and the vertex is at (0, 1/4).

Please note that the graph is a rough approximation and may not be perfectly accurate.