Parabola Families Practice Problems & Transformations

Master parabola transformations with practice problems covering y=x², y=x²+c, y=(x-p)², and y=(x-p)²+k. Learn horizontal and vertical shifts step-by-step.

📚Master Parabola Family Transformations Through Practice
  • Identify vertex and axis of symmetry for basic parabola y=x²
  • Solve vertical shift problems using y=x²+c transformations
  • Determine horizontal shifts in y=(x-p)² parabola families
  • Combine horizontal and vertical transformations in y=(x-p)²+k
  • Graph parabola families with different values of a, p, and k
  • Analyze increasing and decreasing intervals of transformed parabolas

Understanding Parabola Families

Complete explanation with examples

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.

Detailed explanation

Practice Parabola Families

Test your knowledge with 36 quizzes

What is the value of X for the function?

\( y=x^2 \)

of the point \( y=36 \)?

Examples with solutions for Parabola Families

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

Find the ascending area of the function

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

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

Video Solution
Exercise #3

One function

y=6x2 y=6x^2

to the corresponding graph:

1234

Step-by-Step Solution

The function given is y=6x2 y = 6x^2 . 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 x2 x^2 is 6 6 . A positive coefficient indicates that the parabola opens upwards. The value of 6 6 means the parabola is relatively narrow, as it is stretched vertically compared to the standard y=x2 y = x^2 .

To identify the corresponding graph:

  • Recognize that a function of the form y=ax2 y = ax^2 with a>1 a > 1 indicates a narrower parabola.
  • Out of the given graphs, we should look for an upward-opening narrow parabola.

Upon examining each graph, you find that option 2 shows a parabola that is narrower than the standard parabola y=x2 y = x^2 and opens upwards distinctly, matching our function y=6x2 y = 6x^2 .

Therefore, the correct graph for the function y=6x2 y = 6x^2 is option 2.

Answer:

2

Video Solution
Exercise #4

One function

y=2x23 y=-2x^2-3

to the corresponding graph:

333333-3-3-3333-3-3-3-3-3-31234

Step-by-Step Solution

To solve this problem, we'll match the given function y=2x23 y = -2x^2 - 3 with its corresponding graph based on specific characteristics:

  • The function y=2x23 y = -2x^2 - 3 is a quadratic equation representing a parabola.
  • Since the coefficient of x2 x^2 is negative, the parabola opens downward.
  • The y-intercept is -3, which means the parabola crosses the y-axis at 3-3.
  • The maximum point (vertex) of the parabola occurs at its axis of symmetry, from which we know it opens downward from that point.

Given these observations, we analyze each graphical option:

  • Graph 1 represents a parabola opening upward, so it does not match.
  • Graph 2 might have an appropriate direction but not the correct intercept.
  • Graph 3 doesn't match key features such as y-intercept and direction.
  • Graph 4 shows a downward opening parabola with its intercept significantly influenced by negative vertical shift, which matches y=2x23 y = -2x^2 - 3 .

Therefore, the function y=2x23 y = -2x^2 - 3 matches with graph option 4.

Answer:

4

Video Solution
Exercise #5

One function

y=6x2 y=-6x^2

to the corresponding graph:

1234

Step-by-Step Solution

To solve this problem, we need to match the function y=6x2 y = -6x^2 with its graph. This function represents a downward-opening parabola with the vertex at the origin (0,0)(0,0). The coefficient 6-6 is negative, confirming it opens downwards, and its large absolute value indicates that the parabola closes towards the axis more sharply than a standard y=x2 y = -x^2 curve.

Let's identify the characteristics of y=6x2 y = -6x^2 :
- The graph is a parabola, opening downwards.
- The vertex is at the origin, (0,0)(0,0).
- Symmetric around the y-axis.
- Its steepness is greater than the standard parabola y=x2 y = -x^2 due to the coefficient 6 -6 .

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 y=6x2 y = -6x^2 is option 4.

Answer:

4

Video Solution

Frequently Asked Questions

What is the basic parabola function and its properties?

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The basic parabola function is y=x² with vertex at (0,0) and axis of symmetry at x=0. It's a minimum function that decreases for x<0 and increases for x>0, with all y-values being positive except at x=0.

How does the parameter c affect the parabola in y=x²+c?

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The parameter c creates a vertical shift of the parabola. If c is positive, the parabola shifts up by c units; if c is negative, it shifts down by |c| units. The vertex moves from (0,0) to (0,c).

Why does y=(x-p)² move the parabola right when p is positive?

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In y=(x-p)², when p is positive, there's a minus sign in the equation. The parabola shifts p units to the right because the function equals zero when x=p, making the new vertex at (p,0).

How do you find the vertex of y=(x-p)²+k?

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The vertex of y=(x-p)²+k is at point (p,k). The value p determines horizontal shift (right if positive, left if negative), and k determines vertical shift (up if positive, down if negative).

What happens when the coefficient a changes in y=ax²?

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When |a| > 1, the parabola becomes narrower (smaller opening). When 0 < |a| < 1, the parabola becomes wider (larger opening). If a is negative, the parabola opens downward instead of upward.

How do you determine if a parabola opens up or down?

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Look at the coefficient of x². If it's positive (like in y=x² or y=2x²), the parabola opens upward (minimum function). If it's negative (like in y=-x²), the parabola opens downward (maximum function).

What are the increasing and decreasing intervals for parabola families?

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For upward-opening parabolas with vertex at (h,k): decreasing interval is x < h, increasing interval is x > h. For downward-opening parabolas: increasing interval is x < h, decreasing interval is x > h.

How do you graph y=(x+3)²-4 step by step?

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Start with basic parabola y=x². Since p=-3 (from x+3), shift 3 units left. Since k=-4, shift 4 units down. The vertex is at (-3,-4) and the parabola opens upward with axis of symmetry at x=-3.

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