Parabola of the Form y=x²+c - Examples, Exercises and Solutions

The function given is $y = 6x^2$. This is a quadratic function, a type of parabola with vertex at the origin (0,0), because there are no additional terms indicating a horizontal or vertical shift.

First, note the coefficient of $x^2$ is $6$. A positive coefficient indicates that the parabola opens upwards. The value of $6$ means the parabola is relatively narrow, as it is stretched vertically compared to the standard $y = x^2$.

To identify the corresponding graph:

• Recognize that a function of the form $y = ax^2$ with $a > 1$ indicates a narrower parabola.
• Out of the given graphs, we should look for an upward-opening narrow parabola.

Upon examining each graph, you find that option 2 shows a parabola that is narrower than the standard parabola $y = x^2$ and opens upwards distinctly, matching our function $y = 6x^2$.

Therefore, the correct graph for the function $y = 6x^2$ is option 2.

To determine the intervals where the function $f(x) = 2x^2$ is increasing, we will analyze the derivative of the function:

Step 1: Differentiate the function.
The derivative of $f(x) = 2x^2$ is $f'(x) = 4x$.

Step 2: Determine where $f'(x) > 0$.
To find the increasing intervals, set $4x > 0$. Solving this inequality, we obtain $x > 0$.

Therefore, the function $f(x) = 2x^2$ is increasing for $x > 0$.

Consequently, the correct answer is the interval where the function is increasing, which is $0 < x$.

To solve the problem of finding the descending area of the function $f(x) = \frac{1}{2}x^2$, we follow these steps:

• Step 1: Calculate the derivative of the given function. The function is $f(x) = \frac{1}{2}x^2$. Differentiating this, we get $f'(x) = \frac{d}{dx}(\frac{1}{2}x^2) = x$.
• Step 2: Determine where the derivative is negative. Since $f'(x) = x$, the derivative is negative when $x < 0$.
• Step 3: Conclude the solution. We find that the function $f(x)$ is decreasing for $x < 0$.

Thus, the descending area (domain where the function is decreasing) for the function $f(x) = \frac{1}{2}x^2$ is $x < 0$.

The correct choice that matches this solution is: $x < 0$.

To solve the problem of identifying which chart represents the function $y = x^2 - 9$, let's analyze the function and its graph:

• The function $y = x^2 - 9$ is a parabola that can be described by the general form $y = x^2 + k$ where $k = -9$.
• It is a standard upward-opening parabola with its vertex located at the point $(0, -9)$. This is because there is no coefficient affecting $x$, so horizontally it is centered at the origin.
• To find the correct graph, we look for one where the bottommost point of the parabola is at $(0, -9)$. This point, known as the vertex, should sit on the y-axis and be the lowest point of the curve due to the upward opening.

After inspecting the charts:

• Chart 4 depicts a parabola opening upwards, with its vertex at $(0, -9)$. This aligns perfectly with the form and properties of our function $y = x^2 - 9$.

Therefore, the chart that represents the function $y = x^2 - 9$ is Choice 4.

To solve this problem, we'll match the given function $y = -2x^2 - 3$ with its corresponding graph based on specific characteristics:

• The function $y = -2x^2 - 3$ is a quadratic equation representing a parabola.
• Since the coefficient of $x^2$ is negative, the parabola opens downward.
• The y-intercept is -3, which means the parabola crosses the y-axis at $-3$.
• The maximum point (vertex) of the parabola occurs at its axis of symmetry, from which we know it opens downward from that point.

Given these observations, we analyze each graphical option:

• Graph 1 represents a parabola opening upward, so it does not match.
• Graph 2 might have an appropriate direction but not the correct intercept.
• Graph 3 doesn't match key features such as y-intercept and direction.
• Graph 4 shows a downward opening parabola with its intercept significantly influenced by negative vertical shift, which matches $y = -2x^2 - 3$.

Therefore, the function $y = -2x^2 - 3$ matches with graph option 4.

To solve this problem, we need to match the function $y = -6x^2$ with its graph. This function represents a downward-opening parabola with the vertex at the origin $(0,0)$. The coefficient $-6$ is negative, confirming it opens downwards, and its large absolute value indicates that the parabola closes towards the axis more sharply than a standard $y = -x^2$ curve.

Let's identify the characteristics of $y = -6x^2$:
- The graph is a parabola, opening downwards.
- The vertex is at the origin, $(0,0)$.
- Symmetric around the y-axis.
- Its steepness is greater than the standard parabola $y = -x^2$ due to the coefficient $-6$.

By analyzing the given graph options, the graph marked as 4 aligns perfectly with these properties: It is centered on the origin, opens downwards, and has an evident steep slope.

Therefore, the correct graph that matches the function $y = -6x^2$ is option 4.

The function given is $y = \frac{x^2}{4} + 2$, which is a quadratic function with a vertex at $(0, 2)$. The function is in the form $y = a(x-h)^2 + k$, where $a = \frac{1}{4}$, $h = 0$, and $k = 2$. This tells us that the parabola opens upwards with its vertex at $(0, 2)$, and it's wider than the standard parabola $y = x^2$ because $\frac{1}{4}$ is less than 1.

To find the correct graph, look for the one featuring a vertex at $(0, 2)$ with an upward opening, and wider spread due to the smaller coefficient. When comparing the graphs, the graph labeled as choice 1 clearly shows these characteristics, indicating the correct match for the function.

Therefore, the solution corresponds to the graph labeled as choice 1.

The solution to the problem is choice 3.

This problem involves determining the algebraic representation of a parabola that was presented graphically. Our goal is to interpret the graph and express it in terms of its equation for a downward-opening parabola.

To solve the problem, follow these steps:

• Step 1: Identify the Parabola's Vertex – According to the diagram, the vertex is positioned at $(0, 1)$, implying that at $x = 0$, the maximum value of $y$ is 1. This indicates that the constant term $c$ in the parabola's equation will be 1.
• Step 2: Determine the Parabola’s Orientation – The given parabola is described as downward-opening. This means the coefficient in front of $x^2$ must be negative. This leads us to the formula $y = -x^2 + c$.
• Step 3: Construct the Equation – With the downward orientation and the vertex point established, the equation becomes $y = -x^2 + 1$ as $c = 1$ from the vertex.

By matching one of the multiple-choice answers with our derived equation, it's clear that choice 2 corresponds to $y = -x^2 + 1$. Thus

Therefore, the algebraic representation of the function is $y = -x^2 + 1$.

To solve for the graph that matches the function $y = -\frac{1}{2}x^2 + 4$, let's analyze the function:

• The function $y = -\frac{1}{2}x^2 + 4$ is a parabola in standard form $y = ax^2 + bx + c$, with $a = -\frac{1}{2}$, $b = 0$, and $c = 4$.
• Because $a = -\frac{1}{2}$ is negative, the parabola opens downwards.
• The vertex of the parabola $y = ax^2 + bx + c$ is at $x = -\frac{b}{2a}$. Here, $b = 0$, so $x = 0$.
• Substituting $x = 0$ back into the equation gives the vertex's y-coordinate: $y = -\frac{1}{2}(0)^2 + 4 = 4$.
• Thus, the vertex is $(0, 4)$.

Now, let's match this to the graphs:

• We are looking for a graph with a vertex at $(0, 4)$ that opens downwards.
• Upon reviewing the graphs in the problem, graph number 1 presents a downward opening parabola with a vertex at the point $(0, 4)$.

Therefore, the graph that corresponds to $y = -\frac{1}{2}x^2 + 4$ is graph 1.

Thus, the solution to the problem is 1.

To solve this problem, we will determine the vertical shift given to the parent function $y = x^2$ to form the observed parabola.

• Identify that the problem involves a vertical translation of the parabola $y = x^2$.

• The function takes the form $y = x^2 + c$, where $c$ indicates the vertical shift.

• From the graph given, it is seen that the vertex of the parabola is situated at $y = -6$ when viewed from the intersection with the y-axis.

• This downward shift corresponds to the constant $c$ being negative, specifically $c = -6$.

• By this observation, the function becomes $y = x^2 - 6$.

Therefore, the solution to the problem is $y = x^2 - 6$, matching choice 2.

In this problem, we are tasked with identifying the algebraic representation of a function given a graphical depiction. Given the problem's indication that we are dealing with parabolas, particularly those of the form $y = x^2 + c$, we need to examine the provided graph for features typical of this family of functions.

The graph structure in the problem suggests a parabolic curve, centered symmetrically, which is indicative of the simplest unmodified parabola, $y = x^2$. The vertex likely lies at the origin, and the parabola opens upwards, a key characteristic of the function $y = x^2$ when the coefficient of $x^2$ is positive and equal to 1.

Upon reviewing the multiple-choice options, the expression that corresponds to this graph is:

• Option 1: $y = x^2$

Therefore, the algebraic representation that corresponds to the function is $y = x^2$.

Let's solve the problem.

Step 1: Calculate the derivative of the function:

The function is $f(x) = -3x^2 + 12$.

The derivative $f'(x)$ is calculated using the power rule:

$f'(x) = \frac{d}{dx}(-3x^2 + 12) = -6x$.

Step 2: Find where the derivative is positive:

To find the interval where the function is increasing, solve the inequality $f'(x) > 0$.

This yields:

$-6x > 0$

Divide both sides by $-6$ (remember to reverse the inequality sign when dividing by a negative):

$x < 0$

Therefore, the function is increasing for $x < 0$.

Thus, the ascending area of the function is $x < 0$.

To determine the ascending area of the function $f(x) = 6x^2 - 12$, we will follow these steps:

• Step 1: Calculate the derivative of the given function.
• Step 2: Set the inequality $f'(x) > 0$ to find the interval where the function is increasing.
• Step 3: Solve the inequality for $x$.

Let's begin with Step 1:
The derivative of $f(x) = 6x^2 - 12$ with respect to $x$ is:

$f'(x) = \frac{d}{dx}(6x^2 - 12) = 12x$.

Step 2: We need to find where $12x > 0$. This requires:

$x > 0$.

Step 3: Therefore, the function $f(x) = 6x^2 - 12$ is increasing when $x > 0$.

Thus, the increasing interval of the function is when $x > 0$.

The solution to the problem is $0 < x$.

To determine where the function $f(x) = x^2$ is positive, we consider the nature of this parabolic graph, which opens upwards.

Step 1: Recognize that the function $f(x) = x^2$ outputs non-negative values for any real number $x$. The graph of this function is a U-shaped parabola.

Step 2: Analyze the values of the function:
- For $x = 0$, $f(0) = 0^2 = 0$.
- For $x \neq 0$, $f(x) = x^2 > 0$, because squaring any non-zero real number results in a positive value.

Therefore, the function is positive for all $x$ except at $x = 0$, where it is zero.

Step 3: Based on the comparison given in the choices, and our calculation, the area of interest is positive for $x \neq 0$.

Thus, the solution to the problem is that the positive area occurs for $x \neq 0$.