The function
the most basic quadratic function:

the most basic quadratic function:
The family of parabolas
The basic quadratic function – with the addition of
In this family, we are given a quadratic function that clearly shows us how the function moves horizontally – how many steps it needs to move right or left.
represents the number of steps the function will move horizontally – right or left.
If is positive – (there is a minus in the equation) – the function will move steps to the right.
If is negative – (and as a result, there is a plus in the equation because minus times minus equals plus) – the function will move steps to the left.
In this quadratic function, we can see a combination of horizontal and vertical shifts:
: Determines the number of steps and the direction the function will move vertically – up or down.
positive – shift up, negative – shift down.
: Determines the number of steps and the direction the function will move horizontally – right or left.
One function
\( y=6x^2 \)
to the corresponding graph:
Find the ascending area of the function
\( f(x)=2x^2 \)
Find the descending area of the function
\( f(x)=\frac{1}{2}x^2 \)
Which chart represents the function \( y=x^2-9 \)?
Find the intersection of the function
\( y=(x-2)^2 \)
With the X
One function
to the corresponding graph:
The function given is . 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 is . A positive coefficient indicates that the parabola opens upwards. The value of means the parabola is relatively narrow, as it is stretched vertically compared to the standard .
To identify the corresponding graph:
Upon examining each graph, you find that option 2 shows a parabola that is narrower than the standard parabola and opens upwards distinctly, matching our function .
Therefore, the correct graph for the function is option 2.
2
Find the ascending area of the function
To determine the intervals where the function is increasing, we will analyze the derivative of the function:
Step 1: Differentiate the function.
The derivative of is .
Step 2: Determine where .
To find the increasing intervals, set . Solving this inequality, we obtain .
Therefore, the function is increasing for .
Consequently, the correct answer is the interval where the function is increasing, which is .
0 < x
Find the descending area of the function
To solve the problem of finding the descending area of the function , we follow these steps:
Thus, the descending area (domain where the function is decreasing) for the function is .
The correct choice that matches this solution is: .
x < 0
Which chart represents the function ?
To solve the problem of identifying which chart represents the function , let's analyze the function and its graph:
After inspecting the charts:
Therefore, the chart that represents the function is Choice 4.
4
Find the intersection of the function
With the X
To solve this problem, we'll find the intersection of the function with the x-axis. The x-axis is characterized by . Hence, we set and solve for .
Let's follow these steps:
Taking the square root of both sides gives .
Adding 2 to both sides results in .
The x-coordinate is , and since it intersects the x-axis, the y-coordinate is .
Therefore, the intersection point of the function with the x-axis is .
The correct choice from the provided options is .
Complete:
The missing value of the function point:
\( f(x)=x^2 \)
\( f(?)=16 \)
What is the value of y for the function?
\( y=x^2 \)
of the point \( x=2 \)?
One function
\( y=-2x^2-3 \)
to the corresponding graph:
One function
\( y=-6x^2 \)
to the corresponding graph:
Find the intersection of the function
\( y=(x+4)^2 \)
With the Y
Complete:
The missing value of the function point:
To solve this problem, we'll follow these steps:
Now, let's work through each step:
Step 1: We start with the equation given by the function . We know , so we can write:
Step 2: To solve for , we take the square root of both sides of the equation:
Step 3: Solve for :
The square root of 16 is 4, so:
or
This gives us the two solutions: and .
Step 4: Compare these solutions to the answer choices. The correct choice is:
and
Therefore, the solution to the problem is and .
What is the value of y for the function?
of the point ?
To solve this problem, we'll follow these steps:
Now, let's work through each step:
Step 1: The given equation is . We need to substitute into this equation.
Step 2: Substitute to get . Calculate .
Therefore, the value of when is .
Hence, the solution to the problem is .
One function
to the corresponding graph:
To solve this problem, we'll match the given function with its corresponding graph based on specific characteristics:
Given these observations, we analyze each graphical option:
Therefore, the function matches with graph option 4.
4
One function
to the corresponding graph:
To solve this problem, we need to match the function with its graph. This function represents a downward-opening parabola with the vertex at the origin . The coefficient is negative, confirming it opens downwards, and its large absolute value indicates that the parabola closes towards the axis more sharply than a standard curve.
Let's identify the characteristics of :
- The graph is a parabola, opening downwards.
- The vertex is at the origin, .
- Symmetric around the y-axis.
- Its steepness is greater than the standard parabola due to the coefficient .
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 is option 4.
4
Find the intersection of the function
With the Y
To solve this problem, we will find the intersection of the function with the Y-axis by following these steps:
Now, let's solve the problem:
Step 1: Identify the Y-axis intersection by setting .
Step 2: Substitute into the function:
Step 3: The intersection point on the Y-axis is .
Therefore, the solution to the problem is .
What is the positive domain of the function below?
\( y=(x-2)^2 \)
One function
\( y=\frac{x^2}{4}+2 \)
to the corresponding graph:
Find the negative area of the function
\( y=(x+2)^2 \)
Find the positive area of the function
\( y=(x+6)^2 \)
Given the function:
\( y=x^2 \)
Is there a point for ? \( y=16 \)?
What is the positive domain of the function below?
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.
all x,
One function
to the corresponding graph:
The function given is , which is a quadratic function with a vertex at . The function is in the form , where , , and . This tells us that the parabola opens upwards with its vertex at , and it's wider than the standard parabola because is less than 1.
To find the correct graph, look for the one featuring a vertex at 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.
1
Find the negative area of the function
The function describes a parabola that opens upwards and has its vertex at . Since the equation involves a perfect square, it yields only non-negative values for all 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.
There is no
Find the positive area of the function
The given function is . This function is a parabola open upwards with a vertex at .
The expression signifies the square of a number, which is always non-negative for all real numbers . This means .
To find when the area under the curve is positive, solve for when . The square of any non-zero number is positive. Therefore, we require:
.
Simplifying this equation, we find:
Conclusively, the positive area of this parabola exists at all points except precisely at , where the function equals zero.
Looking at the multiple-choice options, the correct answer that aligns with our solution is:
.
Given the function:
Is there a point for ? ?
The problem asks us to find an such that in the function , the value of becomes 16. To do this, we'll substitute into the equation and solve for .
1. Start with the equation of the function:
2. Substitute into the equation:
3. Solve for :
4. Identify the points on the function for these values of :
Among the given options, the point we find in the choices is:
Therefore, the correct answer is the choice that corresponds with this point: