Lesson Objectives

- Learn how to graph a horizontal parabola

## How to Graph a Horizontal Parabola

In this lesson, we want to learn how to graph a horizontal parabola. Previously, we learned how to graph a vertical parabola. We saw the two forms of a vertical parabola:

Example #1: Sketch the graph of each. $$x=2y^2 - 4y + 1$$ Let's write our equation in vertex form. $$x=a(y - k)^2 + h$$ $$k=-b/2a$$ $$b=-4, a=2$$ $$k=-\frac{-4}{4}=-(-1)=1$$ $$x=2(y - 1)^2 + h$$ To find h, plug in a 1 for y: $$h=2(1)^2 - 4(1) + 1$$ $$h=2 - 4 + 1=-1$$ $$x=2(y - 1)^2 - 1$$ So our vertex occurs at (-1, 1). Now we can plot this point: We can find additional points using the step pattern. To do this, multiply a, which is 2, by the pattern:{1, 3, 5, 7, ...}$$1 \cdot 2=2$$ $$3 \cdot 2=6$$ We really only need two additional points. So this pattern tells us if we start at the vertex and move up by one unit, we go 2 units right. $$(-1 + 2, 1 + 1)=(1, 2)$$ From this point, we can move one more unit up, and then go 6 units right. $$(1 + 6, 2 + 1)=(7, 3)$$ Now, we can reflect these points across the axis of symmetry (y = 1) and get the points: $$(1,0),(7,-1)$$

### Standard Form of a Vertical Parabola

$$f(x)=ax^2 + bx + c$$ We learned that we could use the vertex formula to find the vertex form of our parabola.### Vertex Form of a Vertical Parabola

$$f(x)=a(x - h)^2 + k$$ Recall that h is found as -b/2a, and k is found as f(h).### Standard Form of a Horizontal Parabola

$$x=ay^2 + by + c$$ We see that x and y have switched roles here. We can also find the vertex form.### Vertex Form of a Horizontal Parabola

$$x=a(y - k)^2 + h$$ Since x and y have switched roles, -b/2a will give us the k-value. Then we can plug k in for y and find our h-value. Once we have our vertex (h, k), we can plot this and gain additional points using our step pattern. Recall that our step pattern includes multiplying our value for a by the pattern:{1, 3, 5, 7,...}. Now, as we move one unit up, we move horizontally by the number of units given in the pattern. We can then reflect across the axis of symmetry (y = k) to gain additional points. Let's look at an example.Example #1: Sketch the graph of each. $$x=2y^2 - 4y + 1$$ Let's write our equation in vertex form. $$x=a(y - k)^2 + h$$ $$k=-b/2a$$ $$b=-4, a=2$$ $$k=-\frac{-4}{4}=-(-1)=1$$ $$x=2(y - 1)^2 + h$$ To find h, plug in a 1 for y: $$h=2(1)^2 - 4(1) + 1$$ $$h=2 - 4 + 1=-1$$ $$x=2(y - 1)^2 - 1$$ So our vertex occurs at (-1, 1). Now we can plot this point: We can find additional points using the step pattern. To do this, multiply a, which is 2, by the pattern:{1, 3, 5, 7, ...}$$1 \cdot 2=2$$ $$3 \cdot 2=6$$ We really only need two additional points. So this pattern tells us if we start at the vertex and move up by one unit, we go 2 units right. $$(-1 + 2, 1 + 1)=(1, 2)$$ From this point, we can move one more unit up, and then go 6 units right. $$(1 + 6, 2 + 1)=(7, 3)$$ Now, we can reflect these points across the axis of symmetry (y = 1) and get the points: $$(1,0),(7,-1)$$

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