Parabola y=x²+c Practice Problems - Vertical Shifts

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.

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.