The functions y=x²

🏆Practice parabola of the form y=x²

The functions (y=x2,y=x2,y=ax2)(y=x^2,y=-x^2,y=ax^2 )

Y=X2Y=X^2
A- The basic functions   Y=X²

Properties of the function:

The most basic quadratic function b=0b=0,c=0c=0
Minimum, happy face function, its vertex is (0,0)(0,0)
The axis of symmetry of this function is X=0X=0.
The function's interval of increase: X>0X>0
The function's interval of decrease: X<0X<0
Set of positivity: Every XX except 00.
Set of negativity: None. The entire parabola is above the axisXX.

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Test yourself on parabola of the form y=x²!

What is the value of y for the function?

\( y=x^2 \)

of the point \( x=2 \)?

Practice more now
y=x2y=-x^2
A1 - Basic functions Y=-X²

Properties of the Function

The most basic quadratic function a=1a=-1,b=0b=0 ,c=0c=0
Maximum, sad face function, its vertex is (0,0)(0,0)
The axis of symmetry of this function is X=0X=0.
The interval of increase of the function: X<0X<0
The interval of decrease of the function: X>0X>0
Set of positivity: None. The entire parabola is below the axisXX.
Set of negativity: All XX except for X=0X=0


y=ax2y=ax^2

A2 - Basic functions   Y=ax²

Properties of the function:
The quadratic function 
anynumber=aany number = a,b=0b=0,c=0c=0

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

As aa  increases, the parabola will have a smaller opening - closer to its axis of symmetry.
As aa  decreases, the parabola will have a larger opening - further from its axis of symmetry. 


Examples and exercises with solutions for the functions y=x²

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

What is the value of X for the function?

y=x2 y=x^2

of the point y=36 y=36 ?

Video Solution

Step-by-Step Solution

To solve the problem, we will proceed with the following steps:

  • Identify the provided equation and condition.
  • Apply the square root property to solve the equation.
  • Verify the solution with the given choices.

Step-by-step solution:

Step 1: Substitute y=36 y = 36 into the equation y=x2 y = x^2 , which gives:

x2=36 x^2 = 36

Step 2: Solve for x x by taking the square root of both sides. Using the square root property, we have:

x=±36 x = \pm \sqrt{36}

Since the square root of 36 is 6, we find that:

x=±6 x = \pm 6

Therefore, the solutions to the equation are x=6 x = 6 and x=6 x = -6 .

Thus, the value of x x for y=36 y = 36 in the function y=x2 y = x^2 is x=±6 x = \pm 6 .

Answer

x=±6 x=\pm6

Exercise #4

What is the value of y for the function?

y=x2 y=x^2

of the point x=6 x=6 ?

Video Solution

Step-by-Step Solution

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

  • Step 1: Identify the value given for x x .
  • Step 2: Substitute the given x x value into the function.
  • Step 3: Calculate the resulting value for y y .

Now, let's work through each step:
Step 1: The problem states that x=6 x = 6 .
Step 2: Using the function y=x2 y = x^2 , we substitute x=6 x = 6 .
Step 3: Perform the calculation: y=62 y = 6^2 .

Calculating 62 6^2 , we get 36 36 .
Therefore, for the function y=x2 y = x^2 , when x=6 x = 6 , the value of y y is y=36 y = 36 .

Answer

y=36 y=36

Exercise #5

Complete:

The missing value of the function point:

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

f(?)=25 f(?)=25

Video Solution

Step-by-Step Solution

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

  • Step 1: Set up the equation based on the function f(x)=x2 f(x)=x^2 for f(?)=25 f(?)=25 .
  • Step 2: Solve for x x by applying the square root operation.

Now, let's work through each step:
Step 1: We start with the equation x2=25 x^2 = 25 derived from f(x)=25 f(x) = 25 .
Step 2: To solve for x x , we take the square root of both sides:

x=±25 x = \pm \sqrt{25}

Calculating the square root gives us x=±5 x = \pm 5 . However, we are looking for a specific point that fits one of the answer choices:
Therefore, the solution based on the choices provided is x=5 x = 5 .

Concluding, the missing value of the function point is f(5) f(5) , which coincides with choice 1.

Answer

f(5) f(5)

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