Parabola of the Form y=(x-p)² - Examples, Exercises and Solutions

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.

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:

$x \neq -6$.

To solve this problem, we'll follow these steps:

• Step 1: Identify where the quadratic expression $(x+5)^2$ equals zero, as this determines when $y=0$.
• Step 2: Solve the equation $(x+5)^2 = 0$ to find values of $x$.
• Step 3: Determine the values of $x$ where $(x+5)^2 > 0$.

Now, let's work through each step:

Step 1: We need to analyze the expression $(x+5)^2$.

Step 2: Solve $(x+5)^2 = 0$.
The equation simplifies to:

$(x+5) = 0$
Solve for $x$:
$x = -5$

Step 3: Determine $x$ values where $(x+5)^2$ is positive.
The expression $(x+5)^2$ is positive for any $x \neq -5$, because the square of a non-zero real number is always positive.

Therefore, the quadratic $(x+5)^2$ is positive for $x \neq -5$, meaning the positive area applies for all $x$ except $x = -5$.

The correct choice is: For each $x \neq -5$.

Therefore, the solution to the problem is For each $x \neq 5$.

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.

To find the negative area of the given parabola, we need to determine where the function $y = (x + 3)^2 - 1$ is below the x-axis. This corresponds to finding when the parabola is negative.

First, let's set the equation $y = (x + 3)^2 - 1$ equal to zero and solve for $x$ to find the roots:

• Set $(x + 3)^2 - 1 = 0$.

• This simplifies to $(x + 3)^2 = 1$.

• Taking square roots gives $x + 3 = \pm 1$.

• Thus, $x + 3 = 1$ gives $x = -2$, and $x + 3 = -1$ gives $x = -4$.

The roots are $x = -4$ and $x = -2$. The parabola opens upwards since the coefficient of $x^2$ is positive. Therefore, it is negative (below the x-axis) between these roots.

To verify, choose a test point between the roots, say $x = -3$:

• Plug into the equation: $y = ((-3) + 3)^2 - 1 = 0 - 1 = -1$, which is negative.

Therefore, the function is negative on the interval -4 < x < -2 .

The correct answer is -4 < x < -2 .

To solve this problem, we need to determine whether substituting $x = 0$ gives us the points of intersection with the x-axis for the parabola $y = (x - p)^2$.

To find the x-intercepts of any function, we set $y = 0$ because the x-intercepts occur where the curve meets the x-axis, which implies a zero output or function value:

• Start with the equation: $y = (x - p)^2$.
• Set $y = 0$ for the x-intercept: $(x - p)^2 = 0$.
• Solving this gives $(x - p) = 0$, which simplifies to $x = p$.
• Thus, the x-intercept happens at the point where $x = p$, not where $x = 0$.

Therefore, substituting $x = 0$ does not provide the x-intercepts. Instead, it provides the y-intercept. Therefore, the statement given in the problem is False.

Thus, the solution to the problem is False.

To determine if the given statement is true, consider the equation of the parabola $y = (x - p)^2$. The y-intercept occurs where the parabola crosses the y-axis, which is at $x = 0$.

Step 1: Substitute $x = 0$ into the equation:

$y = (0 - p)^2 = p^2$

Step 2: Calculate the y-intercept:

The y-intercept of the parabola is $y = p^2$.

Conclusion: The statement "To find the y-axis intercept, you substitute $x = 0$ into the equation and solve for $y$" is indeed True, as applying this method correctly determined the y-intercept for the given form of a parabola. Therefore, the answer to the problem is True.

To solve this problem, let's go through the process of elimination to find the graph corresponding to $y = x^2$.

The function $y = x^2$ is an upward-opening parabola with its vertex located at the origin point (0, 0). It is symmetric about the y-axis.

Based on our problem statements or diagrams, the given function will match with chart '2'. This chart will depict an upward-facing parabolic shape with no horizontal or vertical shifts.

Therefore, the solution to the problem is the chart labeled 2.

To solve this problem, we should follow these steps:

• Step 1: Understand the characteristics of the function $y = -x^2$. This is a downward-opening parabola with its vertex at (0,0).
• Step 2: Compare these characteristics against the provided graphs to find a match.
• Step 3: Analyze each graph to identify the one that matches these characteristics. Specifically, a parabola that opens downwards with a vertex at the origin will be our match.

Let's go through these steps:
- The function $y = -x^2$ opens downward because of the negative coefficient and is centered at the origin. This gives the parabola a vertex at (0, 0).

Upon reviewing the provided graphs, option 2 corresponds to this function, as it depicts a downward-opening parabola with its vertex at the origin (0,0).

Therefore, the solution to the problem is choice 2.

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 identify where the parabolic function $y = (x-5)^2$ is decreasing.

• Step 1: Recognize the vertex of the parabola. The equation $y = (x-5)^2$ indicates the vertex is at $(5, 0)$.
• Step 2: Understand the direction of the parabola. This parabola opens upward as it is in the form $y = (x-p)^2$.
• Step 3: Determine the decreasing interval. For an upward-opening parabola, it decreases on the left side of the vertex.

From the vertex form of the parabola, we can conclude that the function decreases when $x < 5$.

Therefore, the solution to the problem is $x < 5$.