Family of Parabolas y=x²+c: Vertical Shift

🏆Practice parabola of the form y=x²+c

Family of Parabolas y=x2+c y=x²+c : Vertical Shift

The basic quadratic function y=x2y=x^2 with the addition of CC yields the function y=x2+cy=x^2+c
The meaning of CC is the vertical shift of the function upwards or downwards.
If CC is positive: the function will rise by the number of steps shown in CC.
If CC is negative: the function will descend by the number of steps shown in CC

Start practice

Test yourself on parabola of the form y=x²+c!

einstein

Find the ascending area of the function

\( f(x)=2x^2 \)

Practice more now

Let's look at an example

y=x2+3y=x^2+3
The function will rise three steps.

1 - Vertical shift

Additionally, we can see thatCC marks the intersection point on the YY axis.


Examples and exercises with solutions from the family of parabolas y=x²+c

Exercise #1

One function

y=6x2 y=6x^2

to the corresponding graph:

1234

Video Solution

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

Exercise #2

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 #3

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 #4

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

Exercise #5

One function

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

to the corresponding graph:

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

Video Solution

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

Join Over 30,000 Students Excelling in Math!
Endless Practice, Expert Guidance - Elevate Your Math Skills Today
Test your knowledge
Start practice