Parabola y=(x-p)² Practice Problems with Solutions

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 $(0, 16)$.

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.

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.

The given function is $y = (x+6)^2$. This function is a parabola open upwards with a vertex at $x = -6$.

The expression $(x+6)^2$ signifies the square of a number, which is always non-negative for all real numbers $x$. This means $(x+6)^2 \geq 0$.

To find when the area under the curve is positive, solve for when $(x+6)^2 > 0$. The square of any non-zero number is positive. Therefore, we require:

$(x+6) \neq 0$.

Simplifying this equation, we find:

• $(x+6) = 0$ when $x = -6$.
• Hence, $(x+6)^2$ is positive wherever $x \neq -6$.

Conclusively, the positive area of this parabola exists at all points except precisely at $x = -6$, where the function equals zero.

Looking at the multiple-choice options, the correct answer that aligns with our solution is:

$x \neq -6$.