Parabola of the Form y=(x-p)²: Match functions to the appropriate graphs

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 solve this problem, we'll proceed as follows:

• Step 1: Identify the vertex of the function $y=(x+3)^2$.
• Step 2: Determine the direction the parabola opens.
• Step 3: Compare the features of each choice's graph to the characteristics identified.

Let's analyze the function $y=(x+3)^2$:

Step 1: The vertex of the function $y=(x+3)^2$ is at $(p, 0)$. Since $p = -3$, the vertex is at the point $(-3, 0)$.

Step 2: The function is of the form $y=(x+3)^2$, which opens upwards because the coefficient of $(x+3)^2$ is positive.

Step 3: By comparing graphs, we select the one where the parabola has a vertex at $(-3, 0)$ and opens upwards. Looking at the provided choices, choice 4 has a graph with a vertex at $(-3, 0)$ and is consistent with the function opening upwards.

Therefore, the correct graph corresponding to the function $y=(x+3)^2$ is choice 4.

The provided function is $y = (x + 1)^2$, a parabola opening upwards. The vertex of this parabola is at $x = -1$. We need to find a graph where the turning point (vertex) is at $x = -1$.

To solve this, closely examine the given graphs to identify the one where the parabola's vertex appears at $x = -1$ on the horizontal axis and reflects a symmetry around this axis indicating a minimum point at $y = 0$.

Given the choices in the chart:

• Choice 1 depicts a different vertex or orientation.
• Choice 2 also shows a different shift along the x-axis.
• Choice 3 is correctly showing that the vertex occurs at $x = -1$ with the appropriate configuration indicated by the problem’s prediction.
• Choice 4 does not match the needed features.

Therefore, the graph that matches the equation $y = (x + 1)^2$ is Choice 3, as it is the only graph with the correct vertex point at $(-1, 0)$.

Consequently, the solution to this problem is Choice 3.

To solve this problem, we'll determine which graph represents the line given by $y=(x+2)^2$.

• Step 1: The y-intercept for the function $y = x + 2$ is at point $(0, 2)$.

• Step 2: Another point can be found by substituting $x = -4$, giving $y = -4 + 2 = -2$, so point $(-4, -2)$.

• Step 3: Based on these points, we identify a slope of $\frac{2 - (-2)}{0 - (-4)} = 1$.

• Step 4: Check each graph to find the one that includes these details: y-intercept at 2 and another point at $(-4, -2)$.

Upon examining each option, we find that the graph matching these points and features corresponds to choice 3.

Thus, the correct graph is option 3.

We need to match the function $y = -(x-5)^2$ to the correct graph.

• The graph will be a parabola that opens downward because of the negative sign in front.
• The vertex of this parabola is $(5, 0)$ due to the function $y = -(x-5)^2$.

Let’s analyze the characteristics of the graph:

• The parabola is concave down (opens downward) since the square term is negative.
• The vertex point of our equation must be located at $(5, 0)$.
• This point indicates that the graph's highest point is at $x = 5$.

Reviewing the options given in the chart, Option 1 correctly shows the vertex of the parabola at point $(5, 0)$, and it opens downward, as expected from a negative quadratic function.

The graph in accordance with the given function is option 1.

The problem involves matching a given function $y = -(x-4)^2$ with its corresponding graph from multiple choices.

First, let's analyze the function:

• The function is in vertex form $y = a(x-h)^2 + k$, where the vertex is $(h, k)$.
• Here, $a = -1$, $h = 4$, and $k = 0$, so the vertex is $(4, 0)$.
• The negative sign indicates the parabola opens downward.

To match the function with the correct graph:

• Identify that the vertex of the parabola is at $(4, 0)$.
• The parabola is symmetric around the line $x = 4$, opening downwards.

Upon examining the choices, Option 1 clearly shows a parabola with a vertex at $(4, 0)$ opening downward. This matches perfectly with the function $y = -(x-4)^2$.

Therefore, the correct answer to the problem is 1.

The function $y = -(x+2)^2$ represents a downward-opening parabola with the vertex at $(-2, 0)$. This transformation involves a horizontal shift to the left by 2 units from the origin.

• The vertex form of a quadratic function is $y = a(x-h)^2 + k$, where $(h, k)$ represents the vertex.
• In our function, $h = -2$ and $k = 0$, so the vertex is $(-2, 0)$.
• The coefficient of $-1$ indicates that the parabola opens downward.
• Since the vertex $(-2, 0)$ and the downward opening are correctly depicted in graph 3, this aligns with our function.

Therefore, after comparing the characteristics of the function with the given graphs, the corresponding graph for this function is option 3.