Graphing Horizontal Parabolas Lesson

In this lesson, we will learn about horizontal parabolas, which are parabolas that open left or right. We previously learned about graphing vertical parabolas, which are parabolas that open up or down.

Vertex Form of a Vertical Parabola

$$f(x) = a(x - h)^2 + k$$ Recall that the vertex form of a vertical parabola allows us to quickly find the vertex and the axis of symmetry.
Vertex: $$(h, k)$$ Axis of Symmetry: $$x = h$$ The parabola opens up if a > 0 and down if a < 0.

Vertex Form of a Horizontal Parabola

To find the vertex form of a horizontal parabola, we will start with our vertex form of a vertical parabola. $$f(x) = a(x - h)^2 + k$$ 1) Replace f(x) with y: $$y = a(x - h)^2 + k$$ 2) Subtract k away from each side: $$y - k = a(x - h)^2$$ 3) Swap y - k and x - h: $$x - h = a(y - k)^2$$ You might also see this in an alternative form, which is found by simply adding h to both sides. $$x = a(y - k)^2 + h$$

• The vertex still occurs at (h, k)
• The parabola opens right if a > 0
• The parabola opens left if a < 0
• The line y = k is the axis of symmetry

Let's look at an example.
Example #1: Find the vertex form. $$x = 4y^2 - 16y + 19$$ To find the vertex form, we will complete the square.
1) Subtract 19 away from each side of the equation: $$x - 19 = 4y^2 - 16y$$ 2) Factor out 4 from the right side: $$x - 19 = 4(y^2 - 4y)$$ 3) Complete the square: $$x - 19 = 4(y^2 - 4y + 4 - 4)$$ $$x - 19 = 4(y^2 - 4y + 4) - 16$$ $$x - 3 = 4(y^2 - 4y + 4)$$ 4) Factor the perfect-square trinomial: $$x - 3 = 4(y - 2)^2$$ Alternative Form: $$x = 4(y - 2)^2 + 3$$ Vertex: $$(3, 2)$$ Axis of Symmetry: $$y = 2$$ Desmos Link for More Detail

Vertex Formula for a Vertical Parabola

Instead of completing the square, which is quite tedious, we can use a simple formula to find the vertex. Recall from our lesson on the vertex form of a parabola that for a vertical parabola: $$f(x) = ax^2 + bx + c, a ≠ 0$$ $$h = -\frac{b}{2a}, k = c - \frac{b^2}{4a}$$ $$f(x) = a(x - h)^2 + k, a ≠ 0$$

Vertex Formula for a Horizontal Parabola

Let's start with an equation of the form: $$x = ay^2 + by + c, a ≠ 0$$ Since our x and y have swapped roles here, we get -b/2a as the k-value and c - b²/4a as the h-value. More formally, we can derive this by completing the square. $$x = ay^2 + by + c$$ 1) Subtract c away from each side: $$x - c = ay^2 + by$$ 2) Factor out a from the right side: $$x - c = a\left(y^2 + \frac{b}{a}y\right)$$ 3) Complete the square on the right side: $$\left(\frac{b}{a} \cdot \frac{1}{2}\right)^2 = \frac{b^2}{4a^2}$$ $$x - c = a\left(y^2 + \frac{b}{a}y + \frac{b^2}{4a^2} - \frac{b^2}{4a^2}\right)$$ 4) Distribute the a to the final term and simplify: $$x - c = a\left(y^2 + \frac{b}{a}y + \frac{b^2}{4a^2}\right) - a\frac{b^2}{4a^2}$$ $$x - c = a\left(y^2 + \frac{b}{a}y + \frac{b^2}{4a^2}\right) - \frac{b^2}{4a}$$ 5) Add c to both sides: $$x = a\left(y^2 + \frac{b}{a}y + \frac{b^2}{4a^2}\right) + c - \frac{b^2}{4a}$$ 6) Factor the perfect-square trinomial: $$x = a\left(y + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a}$$ 7) Match the vertex form: $$x = a(y - k)^2 + h$$ $$x = a\left(y - \left(-\frac{b}{2a}\right)\right)^2 + c - \frac{b^2}{4a}$$ $$k = -\frac{b}{2a}$$ $$h = c - \frac{b^2}{4a}$$ So our vertex will occur at: $$\left(c- \frac{b^2}{4a}, -\frac{b}{2a}\right)$$ Let's look at an example.
Example #2: Find the vertex form. $$x = 5y^2 + 10y + 6$$ Method 1: Use the shortcut.
1) We will identify a, b, and c: $$a = 5, b = 10, c = 6$$ 2) Plug into the vertex formula: $$k = -\frac{b}{2a}$$ $$k = -\frac{10}{2(5)} = -\frac{10}{10} = -1$$ $$h = c - \frac{b^2}{4a}$$ $$h = 6 - \frac{(10)^2}{4(5)} = 6 - \frac{100}{20} = 6 - 5 = 1$$ 3) Plug the values for h and k into the vertex form: $$x = a(y - k)^2 + h$$ $$x = 5(y - (-1))^2 + 1$$ $$x = 5(y + 1)^2 + 1$$ Method 2: Complete the square. $$x = 5y^2 + 10y + 6$$ 1) Subtract 6 away from each side: $$x - 6 = 5y^2 + 10y$$ 2) Factor out a 5 from the right side: $$x - 6 = 5(y^2 + 2y)$$ 3) Complete the square: $$x - 6 = 5(y^2 + 2y + 1 - 1)$$ $$x - 6 = 5(y^2 + 2y + 1) - 5$$ $$x - 1 = 5(y + 1)^2$$ Alternative Form: $$x = 5(y + 1)^2 + 1$$ Either way, we can see the vertex occurs at (1, -1) and the axis of symmetry is the line y = -1. Desmos Link for More Detail

Graphing Horizontal Parabolas Using a Step Pattern

In our lesson on graphing vertical parabolas, we learned about a simple step pattern. We will modify our procedure to work with a horizontal parabola.

• Write the parabola in vertex form:
  • $$x = a(y - k)^2 + h$$
• 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 up
  • If a is positive, move right by the next number in the step pattern
  • If a is negative, move left 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
• Sketch the graph

Let's look at an example.
Example #3: Sketch the graph. $$x=2y^2 - 4y + 1$$ 1) Write our equation in vertex form: $$a = 2, b = -4, c = 1$$ $$k = -\frac{b}{2a}$$ $$k = -\frac{(-4)}{2(2)} = -\frac{-4}{4} = 1$$ $$h = c - \frac{b^2}{4a}$$ $$h = 1 - \frac{(-4)^2}{4(2)} = 1 - \frac{16}{8} = 1 - 2 = -1$$ Vertex: $$(-1, 1)$$ Axis of Symmetry: $$y = 1$$ Vertex Form: $$x = 2(y - 1)^2 - 1$$ 2) Since a is 2, we will generate the following pattern: $$(2, 6, 10, 14, ...)$$ 3) Let's plot the vertex (-1, 1): 4) From the vertex, we can generate the following points based on the pattern: $$(-1 + 2, 1 + 1) = (1, 2)$$ $$(1 + 6, 2 + 1) = (7, 3)$$ 5) Draw in the axis of symmetry y = 1 and reflect the points (1, 2) and (7, 3) across the axis of symmetry. This will give us the points (1, 0) and (7, -1) respectively: 6) Sketch the graph: Desmos Link for More Detail

Graphing Horizontal Parabolas Using Graphing Transformations

We can also use graphing transformations to graph a horizontal parabola. Let's first get familiar with the parent relation (this is not a function since it does not pass the vertical line test). Desmos Link for More Detail Let's look at an example.
Example #4: Sketch the graph using transformations. $$x = (y - 1)^2 + 1$$ Let's think about each transformation separately. We will begin with the +1 on the outside. $$x = y^2 + 1$$

$$x=y^2$$		$$x=y^2 + 1$$
x	y	x	y
0	0	1	0
1	-1	2	-1
1	1	2	1
4	-2	5	-2
4	2	5	2

From the table above, it is clear that for a given y-value, the x-value has been increased by 1 when compared to our parent relation x = y². This gives us a shift right by 1 unit. This may be a bit tough to understand since we are usually thinking about horizontal transformations as being counterintuitive. In this situation, the x and y have swapped roles and now the horizontal transformations are straightforward, whereas, the vertical transformations will be counterintuitive. Desmos Link for More Detail Now, let's consider the -1 on the inside.

$$x=y^2$$		$$x=(y - 1)^2$$
x	y	x	y
0	0	0	1
1	-1	1	0
1	1	1	2
4	-2	4	-1
4	2	4	3

From the table above, it is clear that when compared to x = y² to obtain the same x-value, the y-value needs to be increased by 1. This is so the -1 on the inside of the parentheses can be undone. Again, since x and y have swapped roles here, we can see this vertical transformation is now the one that is counterintuitive. Desmos Link for More Detail Now that we have observed each transformation separately, let's think about putting the two together. $$x = y^2$$ $$x = (y - 1)^2 + 1$$

$$x=y^2$$		$$x=(y - 1)^2 + 1$$
x	y	x	y
0	0	1	1
1	-1	2	0
1	1	2	2
4	-2	5	-1
4	2	5	3

When compared to x = y², the graph will be shifted 1 unit right and 1 unit up. A point (x, y) on the graph of x = y² will correspond to a point of (x + 1, y + 1) on the graph of x = (y - 1)² + 1. Desmos Link for More Detail

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