Graphing $y^2=x+4$: Your Sideways Parabola Guide

Exploring the Parabola: A Deep Dive into $y^2 = x + 4$

Hey guys! Today, we're diving headfirst into the fascinating world of parabolas with a special focus on the equation $y^2 = x + 4$. This might look a little different from the parabolas you're used to, the ones that go up and down, like $y = x^2$. But trust me, it's just as cool and has some unique properties we'll uncover. We'll break down how to graph it, understand its key features, and even touch on why it's shaped this way. So, grab your graphing calculators or just a piece of paper, and let's get our math on!

First off, let's get this equation into a more familiar form so we can start plotting. We have $y^2 = x + 4$. To make it easier to work with, we can isolate $y$ by taking the square root of both sides. Remember, when you take the square root, you always get a positive and a negative result! So, this gives us $y = \pm \sqrt{x+4}$. This is super important because it tells us that for every valid $x$ value, there are generally two corresponding $y$ values – one positive and one negative. This is what gives our parabola its distinctive sideways orientation.

Now, let's talk about what this means for graphing. Unlike a standard function where each $x$ gives only one $y$, this equation represents a relation where a single $x$ can produce multiple $y$'s. This is why it's not technically a function. The $\pm$ symbol is our clue that the graph will be symmetric across the x-axis. Think about it: if $(x, y)$ is a point on the graph, then $(x, -y)$ must also be on the graph because squaring $-y$ gives you the same result as squaring $y$. This symmetry is a hallmark of parabolas that open sideways.

Let's get our hands dirty with some points, just like in the table you provided. We need to find $x$ values that make the stuff inside the square root, $x+4$, non-negative, because we can't take the square root of a negative number in the real number system. This means $x+4 \ge 0$, which simplifies to $x \ge -4$. So, our graph will only exist for $x$ values of -4 and greater. This is our domain restriction, guys! The smallest $x$-value we can plug in is -4, and this will be the very tip, or vertex, of our parabola.

Let's fill in that table you started. We'll use the equation $y = \pm \sqrt{x+4}$:

• When $x = -2$: $y = \pm \sqrt{-2+4} = \pm \sqrt{2}$. So, $y \approx \pm 1.414$. The points are $(-2, \sqrt{2})$ and $(-2, -\sqrt{2})$. Your table shows $y=1$, which is an approximation, but the exact values are $\sqrt{2}$ and $-\sqrt{2}$.
• When $x = -1$: $y = \pm \sqrt{-1+4} = \pm \sqrt{3}$. So, $y \approx \pm 1.732$. The points are $(-1, \sqrt{3})$ and $(-1, -\sqrt{3})$.
• When $x = 0$: $y = \pm \sqrt{0+4} = \pm \sqrt{4} = \pm 2$. The points are $(0, 2)$ and $(0, -2)$.
• When $x = 5$: $y = \pm \sqrt{5+4} = \pm \sqrt{9} = \pm 3$. The points are $(5, 3)$ and $(5, -3)$.

As you can see, as $x$ increases, the absolute value of $y$ also increases, making our parabola get wider and wider as we move to the right. The vertex, which we found occurs at $x=-4$, would have $y = \pm \sqrt{-4+4} = \pm 0 = 0$. So, the vertex is at $(-4, 0)$.

This vertex is a crucial point. It's the minimum $x$-value our graph reaches. Since the parabola opens to the right, the vertex is the leftmost point.

The equation $y^2 = x + 4$ is the standard form for a parabola that opens horizontally. The general form is $(y-k)^2 = 4p(x-h)$, where $(h, k)$ is the vertex and $p$ is the distance from the vertex to the focus and from the vertex to the directrix. In our case, $y^2 = x + 4$ can be rewritten as $(y-0)^2 = 1(x - (-4))$. Comparing this to the general form, we see that $h = -4$, $k = 0$, and $4p = 1$. This means $p = 1/4$.

What does this $p$ value tell us? The focus is a point $p$ units away from the vertex along the axis of symmetry. Since our parabola opens to the right, the focus will be $p$ units to the right of the vertex. So, the focus is at $(-4 + 1/4, 0)$, which is $(-15/4, 0)$.

The directrix is a line $p$ units away from the vertex on the opposite side of the focus. For a parabola opening right, the directrix is a vertical line to the left of the vertex. So, the directrix is the line $x = -4 - 1/4$, which is $x = -17/4$.

Remember, a parabola is defined as the set of all points that are equidistant from the focus and the directrix. This property is what gives it its shape.

So, to recap, our parabola $y^2 = x + 4$ has:

• Vertex: $(-4, 0)$
• Opens: To the right
• Axis of Symmetry: The x-axis ($y=0$)
• Domain: $x \ge -4$
• Range: All real numbers (since $y$ can be any positive or negative value)
• Focus: $(-15/4, 0)$
• Directrix: $x = -17/4$

Understanding these components helps us visualize and work with parabolas that don't follow the typical $y = ax^2 + bx + c$ pattern. It's all about recognizing how the variables are squared and using that information to determine the orientation and key points. Pretty neat, right? Keep practicing, and you'll be a parabola pro in no time!