Question

1)y=(x+1)², y=12)y=x²+1, y=5​

Answer 1

Ответ:

1) (X-Y)/(X^1/2-Y^1/2)=(X-Y)/((X-Y)^1/2)=(X-Y)^1/2

Answer 2

Y=(x+1)2=x=0

Y=(x2+1=x=2

Answer 3

It seems like you're asking for the intersections between the given quadratic functions and horizontal lines. Let's solve for the points of intersection for each set of equations.

1. For $y = (x+1)^2$ and $y = 12$:

To find the points of intersection, we set the two expressions for $y$ equal to each other and solve for $x$:

$(x+1)^2 = 12$

Expanding the left side, we get:

$x^2 + 2x + 1 = 12$

Rearranging the equation:

$x^2 + 2x - 11 = 0$

We can solve this quadratic equation using the quadratic formula:

$x = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times -11}}{2 \times 1}$

Solving this gives two potential values for $x$. Let's calculate them:

$x_1 = \frac{-2 + \sqrt{48}}{2} \approx 3.73$ $x_2 = \frac{-2 - \sqrt{48}}{2} \approx -5.73$

Now, plug these $x$ values into the equation $y = (x+1)^2$ to get the corresponding $y$ values:

For $x_1 \approx 3.73$: $y_1 = (3.73 + 1)^2 \approx 16.08$

For $x_2 \approx -5.73$: $y_2 = (-5.73 + 1)^2 \approx 36.08$

So, the points of intersection are approximately: $P_1 \approx (3.73, 16.08)$ $P_2 \approx (-5.73, 36.08)$

2. For $y = x^2 + 1$ and $y = 5$:

To find the points of intersection, we set $y$ in both equations equal to each other:

$x^2 + 1 = 5$

Rearranging the equation:

$x^2 = 4$

Taking the square root of both sides, we get two potential values for $x$:

$x_1 = 2$ $x_2 = -2$

Now, plug these $x$ values into the equation $y = x^2 + 1$ to get the corresponding $y$ values:

For $x_1 = 2$: $y_1 = 2^2 + 1 = 5$

For $x_2 = -2$: $y_2 = (-2)^2 + 1 = 5$

So, the points of intersection are: $P_1 = (2, 5)$ $P_2 = (-2, 5)$