5 Easy Steps to Graph Y = 1/2x²

Unveiling the Secrets of Graphing Y = 1/2x²: A Comprehensive Guide

Mastering the art of graphing quadratic equations is essential for unlocking the complexities of algebra. Among these equations, y = 1/2x² stands out as a unique and intriguing function, captivating the minds of mathematicians and students alike. Embarking on this journey, we will delve into the secrets of graphing this parabola, unraveling its distinctive characteristics and exploring its fascinating properties.

The graph of y = 1/2x² is a parabola, an open and smooth curve that resembles an inverted U-shape. Its vertex, the point where the parabola changes direction, lies at the origin (0, 0). The axis of symmetry, a vertical line that divides the parabola into two congruent halves, also coincides with the y-axis. Furthermore, the graph exhibits a positive y-intercept, located at the point (0, 1/2), indicating that the parabola opens upward.

Understanding the behavior of y = 1/2x² is crucial for sketching its graph accurately. Unlike linear equations, which exhibit a constant rate of change, the rate of change for a quadratic equation varies depending on the value of x. As x approaches infinity, the value of y approaches 0, indicating that the parabola approaches the x-axis asymptotically. However, as x approaches negative infinity, the value of y approaches infinity, illustrating the parabola’s boundless upward trajectory.

Plotting the Vertex

To plot the vertex of the parabola y = 1/2x^2, we need to determine its x-coordinate. The x-coordinate of the vertex is given by the formula -b/2a, where a and b are the coefficients of the quadratic equation. In this case, a = 1/2 and b = 0, so the x-coordinate of the vertex is:

x = -0 / (2 * 1/2) = 0

The y-coordinate of the vertex is then found by plugging the x-coordinate back into the equation:

y = 1/2 * (0)^2 = 0

Therefore, the vertex of the parabola y = 1/2x^2 is located at the point (0, 0).

Here is a table summarizing the steps for plotting the vertex of a parabola y = ax^2 + bx + c:

Step	Formula
Find the x-coordinate of the vertex	-b / 2a
Find the y-coordinate of the vertex	Plugging the x-coordinate back into the equation for y

Finding the Focus and Directrix

The focus of a parabola is a point from which rays parallel to the axis of symmetry reflect to form the parabola. The directrix is a line from which rays emanating from the focus reflect to form the parabola. For a parabola of the form y = ax^2 + bx + c, the focus is located at (0, a/4) and the directrix is the line y = -a/4.

Finding the Focus and Directrix for the Equation y = 1/2x^2

The coefficient of x^2 in the given equation, y = 1/2x^2, is 1/2. Therefore, the focus is located at (0, 1/8) and the directrix is the line y = -1/8.

| Feature | Value |
|—|—|
| Focus | (0, 1/8) |
| Directrix | y = -1/8 |

Sketching the Parabola

To graph the parabola y = 1/2x^2, follow these steps:

1. Find the Vertex:

The vertex of a parabola is the point at which it changes direction. For a parabola of the form y = ax^2 + bx + c, the vertex is given by the coordinates (-b/2a, -Δ/4a). In this case, a = 1/2 and b = 0, so the vertex is located at (0, 0).

2. Plot the Vertex:

Plot the vertex (0, 0) on the coordinate plane.

3. Find the Intercepts:

The intercepts are the points where the parabola intersects the x– and y-axes. To find the x-intercepts, set y = 0 and solve for x. Here, we get x^2 = 0, which gives us x = 0. So, the x-intercepts are (0, 0) (the same as the vertex).

To find the y-intercept, set x = 0 and solve for y. Here, we get y = 0, so the y-intercept is also (0, 0).

4. Draw the Symmetry Line:

The parabola is symmetric about the vertical line passing through the vertex. In this case, the symmetry line is the y-axis (x = 0).

5. Plot Additional Points and Sketch the Parabola:

Choose additional values of x on either side of the vertex and calculate the corresponding y-values. Plot these points on the coordinate plane and connect them with a smooth curve. The curve should be symmetric about the y-axis and open upwards (since a is positive).

Table of Points:

x	y
-2	2
-1	1/2
1	1/2
2	2

Determining the Axis of Symmetry

The axis of symmetry for the parabola y = 1/2x^2 is a vertical line that passes through the vertex of the parabola. The vertex is the point where the parabola changes direction. To find the vertex, we can use the formula x = -b/2a, where a and b are the coefficients of the quadratic equation y = ax^2 + bx + c.

In this case, a = 1/2 and b = 0, so the x-coordinate of the vertex is x = -0/2(1/2) = 0.

The y-coordinate of the vertex can be found by substituting the x-coordinate back into the original equation:

y = 1/2(0)^2 = 0

Therefore, the vertex of the parabola is (0, 0).

The axis of symmetry is the vertical line that passes through the vertex, which is x = 0.

Function	Axis of Symmetry
y = 1/2x^2	x = 0

Identifying the Intercepts

The intercepts of a graph are the points where the graph crosses the x- and y-axes. To find the x-intercepts, set y = 0 and solve for x. To find the y-intercept, set x = 0 and solve for y.

Finding the X-intercepts

Set y = 0 and solve for x:

“`
0 = 1/2x^2
x^2 = 0
x = 0
“`

Therefore, the x-intercept is (0, 0).

Finding the Y-intercept

Set x = 0 and solve for y:

“`
y = 1/2(0)^2
y = 0
“`

Therefore, the y-intercept is (0, 0).

Alternative method using the vertex

Since the parabola opens upward and is symmetric about the x-axis, the vertex of the parabola is the midpoint of the x-intercepts. The vertex is also the minimum point of the parabola.

The vertex form of a parabola is:

“`
y = a(x – h)^2 + k
“`

where (h, k) is the vertex.

For the given equation, y = 1/2x^2, the vertex is (0, 0). Therefore, the x- and y-intercepts are both (0, 0).

X-intercept	Y-intercept
(0, 0)	(0, 0)

Calculating the Domain and Range

Domain

The domain of a function is the set of all possible input values for which the function is defined. For the function
$$y = \frac{1}{2}x^2$$, the domain is all real numbers. This is because the function is defined for any real number input, and there are no restrictions on the input values.

Range

The range of a function is the set of all possible output values for the function. For the function
$$y = \frac{1}{2}x^2$$, the range is all non-negative real numbers. This is because the function always produces a non-negative output, and there are no restrictions on the output values.

Plotting Points

To graph the function, we can plot a few points and then connect them with a smooth curve. Here are a few points that we can plot:

x	y
-2	2
-1	0.5
0	0
1	0.5
2	2

Once we have plotted a few points, we can connect them with a smooth curve to get the graph of the function.

Transforming the Equation

To graph y = 1 – 2x², we’ll start by completing the square to rewrite the equation in vertex form.

Completing the Square

Complete the square for the x-term by adding and subtracting the square of half the coefficient, which is (1/2)² = 1/4:

y = 1 – 2x² + 1/4 – 1/4

Factoring and Vertex Form

Factor the expression inside the parentheses as a perfect square trinomial:

y = (1 – 2x² + 1/4) – 1/4

y = (-2)² – 1/4

y = -2(x² – 1/2) – 1/4

Now the equation is in vertex form, y = a(x – h)² + k, where (h, k) represents the vertex:

y = -2(x + 0)² – 1/4

Vertex: (0, -1/4)

Using the Distance Formula

The distance formula can be used to determine the location of points on a graph. The distance formula is as follows:

$$d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$$

where $$d$$ is the distance between points $$(x_1, y_1)$$ and $$(x_2, y_2)$$.

To graph the equation $$y = 1 + 2x^2$$, we can use the distance formula to find the distance between the point $$(0, 1)$$ and any other point on the graph.

For example, to find the distance between the point $$(0, 1)$$ and the point $$(1, 3)$$, we would use the following formula:

$$d = \sqrt{(1 – 0)^2 + (3 – 1)^2}$$

$$d = \sqrt{1 + 4}$$

$$d = \sqrt{5}$$

This means that the distance between the two points is $$\sqrt{5}$$.

We can use the distance formula to find the distance between any two points on the graph. By doing so, we can create a table of distances that can be used to plot the graph. The following table shows the distances between the point $$(0, 1)$$ and several other points on the graph:

$$x$$	$$y$$	$$d$$
0	1	0
1	3	$$\sqrt{5}$$
2	9	4
3	17	8

The distances in the table can be used to plot the graph of the equation $$y = 1 + 2x^2$$. The graph is a parabola that opens up. The vertex of the parabola is at the point $$(0, 1)$$.

Applying Reflections and Translations

To graph the equation y = 1 – 2x², we apply transformations to the parent function f(x) = x².

Translation:

The term -2 shifts the graph 2 units down in the y-direction, so the vertex becomes (0, -1).

Reflection:

The negative coefficient -2 reflects the graph about the x-axis.

Creating a Table

To create a table of values, we substitute various x-values into the equation and solve for the corresponding y-values:

x	y = 1 – 2x²
-2	9
-1	3
0	1
1	3
2	9

Plotting Points and Drawing the Graph

Using the table, we plot the points (0, -1), (-1, 3), (1, 3), and (-2, 9), (2, 9) and draw a smooth curve through them. The resulting parabola opens downwards, has its vertex at (0, -1), and intersects the x-axis at approximately x = -1.41 and x = 1.41.

Analyzing the Shape and Orientation

The equation y = 1/2x^2 represents a parabola, which is a U-shaped curve. The parabola’s orientation depends on the sign of the coefficient of x^2.

Vertex

The vertex of a parabola is the point where it changes direction. For y = 1/2x^2, the vertex is located at the origin (0, 0).

Axis of Symmetry

The axis of symmetry is a vertical line that divides the parabola into two symmetrical halves. For y = 1/2x^2, the axis of symmetry is x = 0.

Opening

The direction in which the parabola opens depends on the sign of the coefficient of x^2. Since the coefficient is positive (1/2), the parabola opens upwards.

Shape

The shape of the parabola is determined by the value of the coefficient of x^2. The larger the absolute value of the coefficient, the narrower the parabola. In this case, the coefficient is 1/2, which results in a moderate width.

Intercepts

The x-intercepts of the parabola are the points where it intersects the x-axis. To find the x-intercepts, set y = 0 and solve for x:

0 = 1/2x^2
x^2 = 0
x = 0

Therefore, the x-intercepts are (0, 0).

The y-intercept is the point where the parabola intersects the y-axis. To find the y-intercept, set x = 0:

y = 1/2(0)^2
y = 0