Parabola Families - 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 find the intersection of the function $y = (x-2)^2$ with the x-axis. The x-axis is characterized by $y = 0$. Hence, we set $(x-2)^2 = 0$ and solve for $x$.

Let's follow these steps:

• Step 1: Set the function equal to zero:

$(x-2)^2 = 0$

• Step 2: Solve the equation for $x$:

Taking the square root of both sides gives $x - 2 = 0$.

Adding 2 to both sides results in $x = 2$.

• Step 3: Find the intersection point coordinates:

The x-coordinate is $x = 2$, and since it intersects the x-axis, the y-coordinate is $y = 0$.

Therefore, the intersection point of the function with the x-axis is $(2, 0)$.

The correct choice from the provided options is $(2, 0)$.

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 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 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.

To solve this problem, we will find the intersection of the function with the Y-axis by following these steps:

• Step 1: Recognize that the intersection with the Y-axis occurs where $x = 0$.
• Step 2: Substitute $x = 0$ into the function $y = (x+4)^2$.
• Step 3: Perform the calculation to find the y-coordinate.

Now, let's solve the problem:

Step 1: Identify the Y-axis intersection by setting $x = 0$.
Step 2: Substitute $x = 0$ into the function:

$y = (0+4)^2 = 4^2 = 16$

Step 3: The intersection point on the Y-axis is $(0, 16)$.

Therefore, the solution to the problem is $(0, 16)$.

In the first step, we place 0 in place of Y:

0 = (x-2)²

We perform a square root:

0=x-2

x=2

And thus we reveal the point

(2, 0)

This is the vertex of the parabola.

Then we decompose the equation into standard form:

y=(x-2)²

y=x²-4x+2

Since the coefficient of x² is positive, we learn that the parabola is a minimum parabola (smiling).

If we plot the parabola, it seems that it is actually positive except for its vertex.

Therefore the domain of positivity is all X, except X≠2.

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 function $y = (x+2)^2$ describes a parabola that opens upwards and has its vertex at $(-2, 0)$. Since the equation involves a perfect square, it yields only non-negative values for all $x$ and always lies on or above the x-axis. Therefore, there is no part of this parabola that crosses below the x-axis, resulting in no "negative area" above or below the x-axis.

In conclusion, the correct answer among the choices is: There is no negative area.

The given function is $y = (x+6)^2$. This function is a parabola open upwards with a vertex at $x = -6$.

The expression $(x+6)^2$ signifies the square of a number, which is always non-negative for all real numbers $x$. This means $(x+6)^2 \geq 0$.

To find when the area under the curve is positive, solve for when $(x+6)^2 > 0$. The square of any non-zero number is positive. Therefore, we require:

$(x+6) \neq 0$.

Simplifying this equation, we find:

• $(x+6) = 0$ when $x = -6$.
• Hence, $(x+6)^2$ is positive wherever $x \neq -6$.

Conclusively, the positive area of this parabola exists at all points except precisely at $x = -6$, where the function equals zero.

Looking at the multiple-choice options, the correct answer that aligns with our solution is:

$x \neq -6$.

The problem asks us to find an $x$ such that in the function $y = x^2$, the value of $y$ becomes 16. To do this, we'll substitute $y = 16$ into the equation and solve for $x$.

1. Start with the equation of the function:

$y = x^2$

2. Substitute $y = 16$ into the equation:

$16 = x^2$

3. Solve $x^2 = 16$ for $x$:

• Take the square root of both sides to solve for $x$:
• $x = \pm \sqrt{16}$
• This gives $x = 4$ or $x = -4$

4. Identify the points on the function for these values of $x$:

• For $x = 4$, the point is $(4, 16)$.
• For $x = -4$, the point is $(-4, 16)$, but this is not provided in the choice list.

Among the given options, the point we find in the choices is:

$(4, 16)$

Therefore, the correct answer is the choice that corresponds with this point:

$(4,16)$