Family of Parabolas y=(x-p)²

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 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 determine the intersection of the function $y = (x-2)^2$ with the y-axis, we set $x = 0$, as the y-axis is defined by all points where $x = 0$.

Substituting $x = 0$ into the equation:

$y = (0 - 2)^2$

Simplifying this expression:

$y = (-2)^2 = 4$

Thus, the intersection point of the function with the y-axis is $(0, 4)$.

Therefore, the solution to the problem is $(0, 4)$.

To find the intersection of the function $y = (x-6)^2$ with the y-axis, we follow these steps:

• Step 1: Identify the known function and approach the problem by setting $x = 0$ since we are looking for the intersection with the y-axis.

• Step 2: Substitute $x = 0$ into the equation $y = (x-6)^2$.

• Step 3: Perform the calculation to find $y$.

Now, execute these steps:
Step 1: We are given the function $y = (x-6)^2$.
Step 2: Substitute $x = 0$ into the equation:
$y = (0-6)^2$
Step 3: Simplify the expression:
$y = (-6)^2 = 36$

The point of intersection with the y-axis is therefore $(0, 36)$.

Thus, the solution to the problem is $(0, 36)$.

In the first step, we place 0 in place of Y:

0 = (x-2)²

We perform a square root:

0=x-2

x=2

And thus we reveal the point

(2, 0)

This is the vertex of the parabola.

Then we decompose the equation into standard form:

y=(x-2)²

y=x²-4x+2

Since the coefficient of x² is positive, we learn that the parabola is a minimum parabola (smiling).

If we plot the parabola, it seems that it is actually positive except for its vertex.

Therefore the domain of positivity is all X, except X≠2.