Family of Parabolas y=(x-p)²

To solve this problem, we'll find the intersection of the function $y = (x-2)^2$ with the x-axis. The x-axis is characterized by $y = 0$. Hence, we set $(x-2)^2 = 0$ and solve for $x$.

Let's follow these steps:

• Step 1: Set the function equal to zero:

$(x-2)^2 = 0$

• Step 2: Solve the equation for $x$:

Taking the square root of both sides gives $x - 2 = 0$.

Adding 2 to both sides results in $x = 2$.

• Step 3: Find the intersection point coordinates:

The x-coordinate is $x = 2$, and since it intersects the x-axis, the y-coordinate is $y = 0$.

Therefore, the intersection point of the function with the x-axis is $(2, 0)$.

The correct choice from the provided options is $(2, 0)$.

To solve this problem, we will find the intersection of the function with the Y-axis by following these steps:

• Step 1: Recognize that the intersection with the Y-axis occurs where $x = 0$.
• Step 2: Substitute $x = 0$ into the function $y = (x+4)^2$.
• Step 3: Perform the calculation to find the y-coordinate.

Now, let's solve the problem:

Step 1: Identify the Y-axis intersection by setting $x = 0$.
Step 2: Substitute $x = 0$ into the function:

$y = (0+4)^2 = 4^2 = 16$

Step 3: The intersection point on the Y-axis is $(0, 16)$.

Therefore, the solution to the problem is $(0, 16)$.

To determine where the function $y = (x-3)^2$ is increasing, consider the following:

• The function is a parabola that opens upwards, centered at the vertex $x = 3$.
• A parabola of the form $y = (x-p)^2$ is increasing on the interval $x > p$.

This means the function $y = (x-3)^2$ begins to increase after the vertex, which is at $x = 3$.

Thus, the area of increase (or ascending area) for this function is when $x > 3$.

Therefore, the correct answer is $3 < x$.

To solve this problem, we will determine when the function $y = (x+5)^2$ is decreasing by using its derivative:

• Step 1: Differentiate $y = (x+5)^2$ with respect to $x$, yielding $\frac{dy}{dx} = 2(x+5)$.
• Step 2: Set the derivative less than zero: $2(x+5) < 0$.
• Step 3: Simplify the inequality to solve for $x$:
  $x + 5 < 0$ implies $x < -5$.

The function is decreasing for values of $x$ that satisfy $x < -5$.

Therefore, the solution is $x < -5$.

The function $y = (x+2)^2$ describes a parabola that opens upwards and has its vertex at $(-2, 0)$. Since the equation involves a perfect square, it yields only non-negative values for all $x$ and always lies on or above the x-axis. Therefore, there is no part of this parabola that crosses below the x-axis, resulting in no "negative area" above or below the x-axis.

In conclusion, the correct answer among the choices is: There is no negative area.