Parabola y=x² Practice Problems with Step-by-Step Solutions

Master parabola functions y=x², y=-x², and y=ax² with interactive practice problems. Learn vertex, axis of symmetry, and graphing techniques through guided exercises.

📚Master Parabola Functions Through Targeted Practice
  • Graph basic parabola functions y=x² and identify key characteristics
  • Determine vertex and axis of symmetry for parabolas of form y=ax²
  • Analyze increasing and decreasing intervals for quadratic functions
  • Find positivity and negativity sets for parabola equations
  • Compare opening width changes when coefficient 'a' varies
  • Solve real-world problems involving basic parabola functions

Understanding Parabola of the form y=x²

Complete explanation with examples

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.

Detailed explanation

Practice Parabola of the form y=x²

Test your knowledge with 6 quizzes

Does the function \( y=x^2 \) pass through the point where y = 36 and x = 6?

Examples with solutions for Parabola of the form y=x²

Step-by-step solutions included
Exercise #1

What is the value of y for the function?

y=x2 y=x^2

of the point x=2 x=2 ?

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

Video Solution
Exercise #2

Complete:

The missing value of the function point:

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

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

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)

Video Solution
Exercise #3

What is the value of y for the function?

y=x2 y=x^2

of the point x=6 x=6 ?

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

Video Solution
Exercise #4

Complete:

The missing value of the function point:

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

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

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)

Video Solution
Exercise #5

Complete:

The missing value of the function point:

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

f(?)=9 f(?)=9

Step-by-Step Solution

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

  • Step 1: Set up the equation x2=9 x^2 = 9 .
  • Step 2: Solve for x x by taking the square root of both sides.
  • Step 3: Choose the correct answer from the given options.

Now, let's work through each step:
Step 1: We set x2=9 x^2 = 9 .
Step 2: Solving for x x , we take the square root of both sides: x=±9 x = \pm \sqrt{9} .
Step 3: This yields two solutions: x=3 x = 3 and x=3 x = -3 .

Comparing these values with the given choices:

  • Choice 1: f(3) f(3) corresponds to x=3 x = 3 .
  • Choice 3: f(3) f(-3) corresponds to x=3 x = -3 .

Both choices f(3) f(3) and f(3) f(-3) are correct, leading us to select the combined choice: Answer A+C.

Answer:

Answer A+C

Video Solution

Frequently Asked Questions

What is the vertex of the parabola y=x²?

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The vertex of y=x² is at the point (0,0). This is the minimum point of the parabola since it opens upward, creating a 'happy face' shape.

How do you find the axis of symmetry for y=x²?

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The axis of symmetry for y=x² is the vertical line x=0 (the y-axis). This line divides the parabola into two identical halves that mirror each other.

What happens to the parabola when you change from y=x² to y=-x²?

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When you change from y=x² to y=-x², the parabola flips upside down. It becomes a 'sad face' function with a maximum at (0,0) instead of a minimum, and opens downward instead of upward.

How does the coefficient 'a' affect the shape of y=ax²?

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The coefficient 'a' controls the width of the parabola opening. When |a| increases, the parabola becomes narrower (closer to the axis of symmetry). When |a| decreases, the parabola becomes wider (further from the axis of symmetry).

Where is the function y=x² increasing and decreasing?

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For y=x², the function decreases on the interval x<0 (left side of vertex) and increases on the interval x>0 (right side of vertex). The vertex at x=0 is the turning point between decreasing and increasing behavior.

What are the positive and negative regions of y=x²?

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For y=x², the function is positive for all x-values except x=0 where y=0. There are no negative y-values since the entire parabola lies above or on the x-axis.

Why is y=x² called a quadratic function?

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y=x² is called a quadratic function because the highest power of the variable x is 2 (squared). It's the most basic form of a quadratic with coefficients a=1, b=0, and c=0 in the standard form y=ax²+bx+c.

How do you graph y=ax² step by step?

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To graph y=ax²: 1) Plot the vertex at (0,0), 2) Draw the axis of symmetry at x=0, 3) Choose x-values and calculate corresponding y-values, 4) Plot points symmetrically on both sides of the axis, 5) Connect points with a smooth U-shaped curve (upward if a>0, downward if a<0).

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