Parabola y=x²+c Practice Problems - Vertical Shifts

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.