The Family of Parabolas

The function y=x2y=x^2

the most basic quadratic function:
y=X2y=X^2

Parabola y=X²

The family of parabolas y=x2+cy=x²+c

The family of parabolas y=x2+cy=x^2+c
The basic quadratic function – with the addition of cc

The family of parabolas y=(xp)2y=(x-p)²

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.
PP represents the number of steps the function will move horizontally – right or left.
If PP is positive – (there is a minus in the equation) – the function will move PP steps to the right.
If PP is negative – (and as a result, there is a plus in the equation because minus times minus equals plus) – the function will move PP steps to the left.

The family of parabolas y=(xp)2+ky=(x-p)²+k

In this quadratic function, we can see a combination of horizontal and vertical shifts:
KK: Determines the number of steps and the direction the function will move vertically – up or down.
KK positive – shift up, KK negative – shift down.
PP: Determines the number of steps and the direction the function will move horizontally – right or left.

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Test yourself on parabola families!

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What is the value of y for the function?

\( y=x^2 \)

of the point \( x=2 \)?

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The Family of Parabolas

Meet the most basic quadratic function:
y=X2y=X^2
In this function:
b=0 ,c=0 ,a=1b=0~, c=0~, a=1

Parabola y=X²

The function is a smiling minimum function and its vertex is - (0,0)(0,0)
The axis of symmetry of this function is X=0X=0.
The increasing interval of the function – all the XX values where the function is increasing are: X>0X>0
The decreasing interval of the function – all the XX values where the function is decreasing are: X<0X<0
Positive interval: all XX except 00 – you can see in the graph that the entire function is above the XX axis
Negative interval: none. The entire function is above the XX axis.
Let's continue to a similar function from the same family:
y=x2y=-x^2
In this function:
a=1, b=0, c=0a=-1 ,~b=0 ,~c=0

Parabola y=-X²

The function is a sad maximum function and its vertex is (0,0)(0,0)
The axis of symmetry of this function is X=0X=0.
Increasing interval: X<0X<0
Decreasing interval: X>0X>0
Positive interval: None. The entire function is below the XX axis.
Negative interval: All XX except X=0X=0
Let's continue with another function from the same family:
y=ax2y=ax^2
In this function:
a=some number, b=0, c=0a=some~number,~b=0,~ c=0

Parabolas y=ax²

Its vertex is- (0,0)(0,0)
The axis of symmetry of this function is X=0X=0.

The larger aa is: the parabola will have a smaller opening – closer to its axis of symmetry.
The smaller aa is: the parabola will have a larger opening – farther from its axis of symmetry.

The function from this family has no horizontal or vertical shift because in every function from this family the vertex is (0,0)(0,0)

The family of parabolas y=x2+cy=x^2+c
The basic quadratic function – with the addition of cc
In this function:
cc – represents the intersection point of the function with the YY axis.
The meaning of cc is a vertical shift up or down of the function from the vertex (0,0)(0,0).
If cc is positive – the function will shift vertically up by the number of steps indicated by cc.
If cc is negative – the function will shift vertically down by the number of steps indicated by cc.

This function describes only vertical shifts up and down according to cc
Let's see an example:
y=4x2+7y=4x^2+7

In this function -
C=7C=7
this means that the intersection point of the function with the YY axis is 77.
And actually – the function moves vertically 77 steps up.

Parabola y=4x²+7

The family of parabolas \(y=(x-p)^2\)

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.
PP represents the number of steps the function will move horizontally – right or left.
If PP is positive – (there is a minus in the equation) – the function will move PP steps to the right.
If PP is negative – (and as a result, there is a plus in the equation because minus times minus equals plus) – the function will move PP steps to the left.
Let's see an example:
   Y=(X6)2    Y=(X-6)^2

This function moves from the vertex (0,0)(0,0),66 steps to the right
Let's see this in the figure:

Parabola Y=(X-6)²

The family of parabolas y=(xp)2+ky=(x-p)^2+k

In this quadratic function, we can see a combination of horizontal and vertical shifts:
KK: Determines the number of steps and the direction the function will move vertically – up or down.
KK positive – shift up, KK negative – shift down.
PP: Determines the number of steps and the direction the function will move horizontally – right or left.

Let's look at an example of combining two translations together:
For example, in the function:
y=(x+2)2+5y=(x+2)^2+5

The changes will be:
According to P=2P=-2 : the parabola will move 22 steps to the left.
According to K=5K=5 : the parabola will move 55 steps up.

Let's see this in the illustration:

Parabola y=(x+2)²+5

We can see that the vertex of the parabola is:
(2,5)(-2,5)

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Examples with solutions for Parabola Families

Exercise #1

What is the value of y for the function?

y=x2 y=x^2

of the point x=2 x=2 ?

Video Solution

Step-by-Step Solution

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

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

Now, let's work through each step:
Step 1: The given equation is y=x2 y = x^2 . We need to substitute x=2 x = 2 into this equation.

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

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

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

Answer

y=4 y=4

Exercise #2

Complete:

The missing value of the function point:

f(x)=x2 f(x)=x^2

f(?)=16 f(?)=16

Video Solution

Step-by-Step Solution

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 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)=x2 f(x) = x^2 . We know f(?)=16 f(?) = 16 , so we can write:

x2=16 x^2 = 16

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

x=±16 x = \pm \sqrt{16}

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

The square root of 16 is 4, so:

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

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

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

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

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

Answer

f(4) f(4) f(4) f(-4)

Exercise #3

Find the ascending area of the function

f(x)=2x2 f(x)=2x^2

Video Solution

Step-by-Step Solution

To determine the intervals where the function f(x)=2x2 f(x) = 2x^2 is increasing, we will analyze the derivative of the function:

Step 1: Differentiate the function.
The derivative of f(x)=2x2 f(x) = 2x^2 is f(x)=4x f'(x) = 4x .

Step 2: Determine where f(x)>0 f'(x) > 0 .
To find the increasing intervals, set 4x>0 4x > 0 . Solving this inequality, we obtain x>0 x > 0 .

Therefore, the function f(x)=2x2 f(x) = 2x^2 is increasing for x>0 x > 0 .

Consequently, the correct answer is the interval where the function is increasing, which is 0<x 0 < x .

Answer

0 < x

Exercise #4

Find the descending area of the function

f(x)=12x2 f(x)=\frac{1}{2}x^2

Video Solution

Step-by-Step Solution

To solve the problem of finding the descending area of the function f(x)=12x2 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)=12x2 f(x) = \frac{1}{2}x^2 . Differentiating this, we get f(x)=ddx(12x2)=x f'(x) = \frac{d}{dx}(\frac{1}{2}x^2) = x .
  • Step 2: Determine where the derivative is negative. Since f(x)=x f'(x) = x , the derivative is negative when x<0 x < 0 .
  • Step 3: Conclude the solution. We find that the function f(x) f(x) is decreasing for x<0 x < 0 .

Thus, the descending area (domain where the function is decreasing) for the function f(x)=12x2 f(x) = \frac{1}{2}x^2 is x<0 x < 0 .

The correct choice that matches this solution is: x<0 x < 0 .

Answer

x < 0

Exercise #5

Which chart represents the function y=x29 y=x^2-9 ?

222333999-9-9-9-1-1-1444-101234

Video Solution

Step-by-Step Solution

To solve the problem of identifying which chart represents the function y=x29 y = x^2 - 9 , let's analyze the function and its graph:

  • The function y=x29 y = x^2 - 9 is a parabola that can be described by the general form y=x2+k y = x^2 + k where k=9 k = -9 .
  • It is a standard upward-opening parabola with its vertex located at the point (0,9)(0, -9). This is because there is no coefficient affecting x 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)(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)(0, -9). This aligns perfectly with the form and properties of our function y=x29 y = x^2 - 9 .

Therefore, the chart that represents the function y=x29 y = x^2 - 9 is Choice 4.

Answer

4

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