Parabola Families Practice Problems & Transformations

To solve this problem, we'll follow these steps:

• Step 1: Substitute the given value of $x$ into the equation.
• Step 2: Perform the calculation to find $y$.

Now, let's work through each step:
Step 1: The given equation is $y = x^2$. We need to substitute $x = 2$ into this equation.

Step 2: Substitute to get $y = (2)^2$. Calculate $2 \times 2 = 4$.

Therefore, the value of $y$ when $x = 2$ is $y = 4$.

Hence, the solution to the problem is $y = 4$.

To solve this problem, we'll follow these steps:

• Step 1: Set up the equation from the function definition.
• Step 2: Solve the equation by taking the square root of both sides.
• Step 3: Identify all possible values for $x$.
• Step 4: Compare with the given answer choices.

Now, let's work through each step:

Step 1: We start with the equation given by the function $f(x) = x^2$. We know $f(?) = 16$, so we can write:

$x^2 = 16$

Step 2: To solve for $x$, we take the square root of both sides of the equation:

$x = \pm \sqrt{16}$

Step 3: Solve for $\sqrt{16}$:

The square root of 16 is 4, so:

$x = 4$ or $x = -4$

This gives us the two solutions: $x = 4$ and $x = -4$.

Step 4: Compare these solutions to the answer choices. The correct choice is:

$f(4)$ and $f(-4)$

Therefore, the solution to the problem is $f(4)$ and $f(-4)$.

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.