Question

Нули функции у=х^2-4, у =(х-5)(х+2) , у=х^2-6хпожалуйста

Answer 1

У=0; х^2 -4 =0; х^2 = 4; х = +-2
у =0; (х -5 ) (х +2 ) =0; х = 5; х = -2
у =0; х^) -6х =0; х (х -6) =0; х =0; х = 6

Answer 2

To find the zeros of each function, we need to determine the values of 'x' that make 'y' equal to zero. In other words, we're looking for the x-values where the graph of the function intersects the x-axis.

1. For the function y = x^2 - 4: To find the zeros, we set y to zero and solve for x: 0 = x^2 - 4

Add 4 to both sides: 4 = x^2

Now, take the square root of both sides (considering both the positive and negative square roots): x = ±√4 x = ±2

The zeros of this function are x = 2 and x = -2.

2. For the function y = (x - 5)(x + 2): Again, set y to zero and solve for x: 0 = (x - 5)(x + 2)

Now, this equation will be true if either (x - 5) equals zero or (x + 2) equals zero:

x - 5 = 0 x = 5

x + 2 = 0 x = -2

The zeros of this function are x = 5 and x = -2.

3. For the function y = x^2 - 6x: Set y to zero and solve for x: 0 = x^2 - 6x

Factor the expression: 0 = x(x - 6)

Once again, this equation will be true if either x or (x - 6) equals zero:

x = 0

x - 6 = 0 x = 6

The zeros of this function are x = 0 and x = 6.

To summarize:

1. For y = x^2 - 4, the zeros are x = 2 and x = -2.
2. For y = (x - 5)(x + 2), the zeros are x = 5 and x = -2.
3. For y = x^2 - 6x, the zeros are x = 0 and x = 6.