Lesson Objectives
• Demonstrate the ability to write a parabola in vertex form
• Learn how to sketch the graph of a parabola using graphing transformations
• Learn how to sketch the graph of a parabola using the step pattern

## How to Sketch the Graph of a Parabola

In this lesson, we will continue to learn about quadratic functions and their graphs, which are known as parabolas. A quadratic function is of the form: $$f(x) = ax^2 + bx + c, a ≠ 0$$ Where a, b, and c are real numbers and a is not zero. In the last lesson, we learned how to find the vertex form of a parabola.

### Vertex Form of a Parabola

$$f(x)=a(x - h)^2 + k$$ As we noted in the previous lesson, some textbooks or other resources will refer to "vertex form" as "standard form". From the vertex form, we immediately know the vertex of the parabola. This occurs at the point (h, k). The vertex of the parabola is the turning point, it will be the lowest point on the graph when it opens upward and the highest point on the graph when it opens downward. If we look at the graph of the squaring function f(x) = x2, the vertex occurs at (0, 0). As we have previously learned, the squaring function is an even function, which means it is symmetric with respect to the y-axis. Not all parabolas will be even but all with be symmetric with respect to a vertical line, known as the axis of symmetry or the axis of the parabola. If a parabola is folded along its axis of symmetry, the two halves will perfectly match. If we return to our vertex form of a parabola: $$f(x)=a(x - h)^2 + k$$
• The vertex occurs at (h, k)
• The axis of symmetry is the vertical line x = h
Let's look at an example.
Example #1: Identify the vertex and axis of symmetry. $$f(x) = (x + 3)^2 - 6$$ First, let's write this in the vertex form. $$f(x) = a(x - h)^2 + k$$ $$f(x) = (x - (-3))^2 + (-6)$$ $$h = -3, k = -6$$ Vertex: $$(-3, -6)$$ Axis of Symmetry: $$x = -3$$ Desmos Link for More Detail
$$f(x) = (x + 3)^2 - 6$$

### Graphing Parabolas Using Graphing Transformations

There are many strategies that can be used to sketch the graph of a parabola. We previously learned how to sketch the graph of a function using graphing transformations. We can graph a parabola by first writing the parabola in vertex form, and then applying the transformations in the correct order. Let's look at an example.
Example #2: Sketch the graph of each. $$g(x) = 2x^2 + 16x + 31$$ First, let's write our parabola in vertex form: $$g(x) = a(x - h)^2 + k$$ $$h = -\frac{b}{2a}, k = f(h)$$ $$a = 2, b = 16$$ $$h = -\frac{16}{2(2)} = -\frac{16}{4} = -4$$ $$k = f(-4) = 2(-4)^2 + 16(-4) + 31$$ $$= 2(16) + 16(-4) + 31$$ $$= 32 - 64 + 31$$ $$= -32 + 31 = -1$$ Vertex Form: $$g(x) = 2(x - (-4))^2 + (-1)$$ Vertex: $$(-4, -1)$$ Axis of Symmetry: $$x = -4$$ $$g(x) = 2(x + 4)^2 - 1$$ Let's think about this function compared to the squaring function. $$f(x) = x^2$$ $$g(x) = 2\cdot f(x + 4) - 1$$ Compared to the graph of f(x), g(x) has been vertically stretched by a factor of 2, shifted left by 4 units, and shifted down by 1 unit. An (x, y) on the graph of f(x), will now be (x - 4, 2y - 1) on the graph of g(x).
x f(x) x g(x)
0 0 -4 -1
-1 1 -5 1
1 1 -3 1
-2 4 -6 7
2 4 -2 7
$$f(x) = x^2$$
$$g(x) = 2(x + 4)^2 - 1$$

### Graphing a Parabola Using the Step Pattern

As we previously mentioned, there are many strategies that can be used to sketch the graph of a parabola. Using graphing transformations is a solid method, however, other methods could be faster, depending on the situation. An alternative approach can be to use a step pattern. Let's look at the graph of f(x) = x2 and observe a simple step pattern: 1, 3, 5, 7,... Starting with the vertex (0, 0), we can move one unit right and one unit up to obtain the point (1, 1). From there, we can move one unit right and three units up to obtain the point (2, 4). If we continue this process, we will obtain the points: (3, 9), (4, 16), (5, 25), and so on. We could then use the y-axis or x = 0, the axis of symmetry to find additional points such as (-1, 1), (-2, 4), (-3, 9), (-4, 16), (-5, 25), and so on. This process will work for any quadratic, however, when the value of a, the coefficient of x2 is not 1, we need to generate the step pattern by multiplying a by each number in the step pattern.

### Using the Step Pattern

• Write the parabola in vertex form:
• $$f(x) = a(x - h)^2 + k$$
• Identify the vertex of the parabola: (h, k)
• Generate the step pattern, multiply a by (1, 3, 5, 7,...)
• Generate points:
• Start at the vertex
• Move one unit right
• If a > 0 Move up by the next number in the step pattern
• If a < 0 Move down by the absolute value of the next number in the step pattern
• Continue this process until you have enough points
• Reflect the points found above across the axis of symmetry to get additional points
Note that this step pattern works best when a is an integer. It can still be used when a is a fraction but it's much more tedious. Let's look at an example.
Example #3: Sketch the graph of each. $$f(x)=2x^2 + 4x - 2$$ First, we will write the parabola in vertex form: $$f(x) = a(x - h)^2 + k$$ $$h = -\frac{b}{2a}, k = f(h)$$ $$a = 2, b = 4$$ $$h = -\frac{4}{2(2)} = -\frac{4}{4} = -1$$ $$k = f(-1) = 2(-1)^2 + 4(-1) - 2$$ $$= 2(1) + 4(-1) - 2$$ $$= 2 - 4 - 2 = -2 - 2 = -4$$ Vertex Form: $$f(x) = 2(x - (-1))^2 + (-4)$$ Vertex: $$(-1, -4)$$ Axis of Symmetry: $$x = -1$$ $$f(x) = 2(x + 1)^2 - 4$$ At this point, we can plot the vertex. Now, we will generate our step pattern, multiply 2 by (1, 3, 5, 7,...), this will give us (2, 6, 10, 14,...). From the vertex, we will move one unit right and two units up. This gives us the point (0, -2). From the point (0, -2), we will move one unit right and six units up. This gives us the point (1, 4). Let's now draw in the axis of symmetry, x = -1, and reflect the points (0, -2) and (1, 4) across the axis of symmetry. This will give us the points (-2, -2) and (-3, 4) respectively. Now that we have these five points, we are ready to sketch the graph of our parabola.

### Write an Equation of a Parabola From a Graph

In some cases, you will be given the graph of a parabola and asked to come up with the equation. Let's look at an example.
Example #4: Find an equation for the given parabola.
Desmos Link for More Detail From the graph, we see the vertex occurs at (-3, -2). Let's plug that into the vertex form. $$f(x) = a(x - h)^2 + k$$ $$h = -3, k = -2$$ $$f(x) = a(x - (-3))^2 + (-2)$$ $$f(x) = a(x + 3)^2 - 2$$ How can we find the value of a? Let's just use another point. We will just pick (-2, 0) but you can use any point you like that is not the vertex. Here, we will replace f(x) with y. $$y = a(x + 3)^2 - 2$$ The point is (-2, 0): $$x = -2, y = 0$$ $$0 = a(-2 + 3)^2 - 2$$ Solve for a: $$a(1) - 2 = 0$$ $$a = 2$$ Now we can replace a with 2 in our function: $$f(x) = 2(x + 3)^2 - 2$$

#### Skills Check:

Example #1

Match the graph to its function.

A
$$f(x) = -(x - 4)^2 - 1$$
B
$$f(x) = -\frac{1}{3}(x + 4)^2 - 1$$
C
$$f(x) = -2(x + 4)^2 - 1$$
D
$$f(x) = -\frac{1}{2}(x + 4)^2 - 1$$
E
$$f(x) = -(x + 4)^2 - 1$$

Example #2

Match the graph to its function.

A
$$f(x) = 2(x + 5)^2 - 2$$
B
$$f(x) = 2(x - 5)^2 - 2$$
C
$$f(x) = 4(x - 5)^2 - 2$$
D
$$f(x) = \frac{1}{2}(x - 5)^2 - 2$$
E
$$f(x) = \frac{1}{4}(x + 5)^2 - 2$$

Example #3

Match the graph to its function.

A
$$f(x) = 3(x - 1)^2 - 4$$
B
$$f(x) = 3(x - 4)^2 - 1$$
C
$$f(x) = 3(x + 1)^2 - 4$$
D
$$f(x) = \frac{1}{3}(x - 1)^2 - 4$$
E
$$f(x) = 3(x + 4)^2 - 1$$