Parabola of the Form y=x²+c: Match functions to the appropriate graphs

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.

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 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.

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.

The solution to the problem is choice 3.

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.

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 need to match the quadratic function $y = 2x^2 + 3$ with one of the graph choices.

First, identify the characteristics of the parabola:

• The standard form of the function is $y = 2x^2 + 3$, which is already in vertex form for vertical shift.
• The parabola opens upwards since the coefficient of $x^2$ is positive ($a = 2$).
• The vertex of the parabola is at $(0, 3)$, not subject to any horizontal shifts. The only transformation from $y = x^2$ is the vertical shift by 3 units up.

Now, assess the graph choices:

• Look for a parabola that is centered on the vertical axis (origin along the x-axis) and opens upwards.
• Among the provided graphs, the one depicting an upright parabola with vertex at $(0, 3)$ should correspond to our function.

The correct choice is graph 3, as it aligns with our function's characteristics: opening upwards, vertex located at $(0, 3)$.

Therefore, the solution to the problem is graph 3.