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
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
of the point x=2?
Step-by-Step Solution
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=x2. We need to substitute x=2 into this equation.
Step 2: Substitute to get y=(2)2. Calculate 2×2=4.
Therefore, the value of y when x=2 is y=4.
Hence, the solution to the problem is y=4.
Answer:
y=4
Video Solution
Exercise #2
Complete:
The missing value of the function point:
f(x)=x2
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.
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. We know f(?)=16, so we can write:
x2=16
Step 2: To solve for x, we take the square root of both sides of the equation:
x=±16
Step 3: Solve for 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).
Answer:
f(4)f(−4)
Video Solution
Exercise #3
What is the value of y for the function?
y=x2
of the point x=6?
Step-by-Step Solution
To solve this problem, we'll follow these steps:
Step 1: Identify the value given for x.
Step 2: Substitute the given x value into the function.
Step 3: Calculate the resulting value for y.
Now, let's work through each step:
Step 1: The problem states that x=6.
Step 2: Using the function y=x2, we substitute x=6.
Step 3: Perform the calculation: y=62.
Calculating 62, we get 36.
Therefore, for the function y=x2, when x=6, the value of y is y=36.
Answer:
y=36
Video Solution
Exercise #4
Given the function:
y=x2
Is there a point for ? y=−2?
Step-by-Step Solution
To solve this problem, we'll follow these steps:
Step 1: Analyze the equation y=x2.
Step 2: Investigate whether a negative y-value is possible.
Now, let's work through each step:
Step 1: The function we have is y=x2. This function is defined for all real numbers and always gives a non-negative value y because squaring a real number cannot result in a negative number.
Step 2: We need to check whether y=−2 is possible by solving x2=−2. In the real number system, no real number x satisfies this equation since the square of any real number is non-negative.
Therefore, there is no real point where y=−2 on the graph of the function y=x2.
Therefore, the solution to the problem is No.
Answer:
No
Video Solution
Exercise #5
Does the function y=x2 pass through the point where y = 36 and x = 6?
Step-by-Step Solution
To determine if the function y=x2 passes through the point (6,36), follow these steps:
Step 1: Identify the given point and function. We have x=6 and we need to check if y=36 when y=x2.
Step 2: Substitute x=6 in the function y=x2: y=62=36.
Step 3: Compare the calculated y value (36) to the given value (36).
Since the calculated value of y is equal to the given value, the function y=x2 indeed passes through the point (6,36).
Therefore, the answer is Yes.
Answer:
Yes
Video Solution
Frequently Asked Questions
Everything you need to know about Parabola of the form y=x²
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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