Is Y A Function Of X? Solving Equations

In mathematics, a function is a fundamental concept that describes a relationship between two sets of values, where each input has exactly one output. When we talk about defining $y$ as a function of $x$, we're essentially asking if, for any given value of $x$, there is only one corresponding value of $y$. Let's dive into some examples and figure out how to determine this!

Understanding Functions: The Vertical Line Test and Beyond

Before we tackle the specific equations, let's solidify our understanding of what it means for $y$ to be a function of $x$. The most intuitive way to think about this is through the Vertical Line Test. If you were to graph an equation, and you could draw a vertical line that intersects the graph at more than one point, then $y$ is not a function of $x$. Why? Because that vertical line represents a single $x$-value, and if it hits the graph multiple times, it means that single $x$-value is associated with multiple $y$-values. A function, by definition, cannot have this ambiguity.

Another way to approach this analytically is to try and solve the equation for $y$. If, after manipulating the equation, you find that for a single value of $x$, you can get more than one possible value for $y$ (often indicated by a $\pm$ sign or a square root that could be positive or negative), then it's not a function. However, if for every $x$, you get one unique $y$, then you've got a function on your hands.

It's also important to remember that the domain and range play a role. A function must be defined for its inputs (domain) and produce unique outputs (range). We're focusing here on the uniqueness of the output for a given input.

Let's get ready to apply these principles to the given equations. We'll go through each one step-by-step, explaining the reasoning behind whether $y$ is a function of $x$ or not. This process will not only help us solve these specific problems but also build a stronger foundation for understanding functions in various mathematical contexts.

Equation 1: $(y+4)^2-36=x$

Let's start with our first equation: $(y+4)^2-36=x$. Our goal here is to see if we can isolate $y$ and determine if each $x$ yields a single $y$. To do this, we'll perform algebraic manipulations.

First, we want to get the term with $y$ by itself. So, we'll add 36 to both sides of the equation:

$(y+4)^2 = x + 36$

Now, to get rid of the square, we'll take the square root of both sides. Crucially, when we take the square root of both sides of an equation, we must remember that there are two possible roots: a positive one and a negative one. This is the key to determining if we have a function.

Taking the square root of both sides gives us:

$y+4 = \pm\sqrt{x+36}$

Finally, to isolate $y$, we subtract 4 from both sides:

$y = -4 \pm\sqrt{x+36}$

Now, let's analyze this result. For any value of $x$ such that $x+36 > 0$ (i.e., $x > -36$), the term $\sqrt{x+36}$ will yield a positive number. Because of the $\pm$ sign, this means we will have two possible values for $y$: one where we add the square root, and one where we subtract it.

For example, if we choose $x=0$, then $y = -4 \pm\sqrt{0+36} = -4 \pm 6$. This gives us two values for $y$: $y = -4 + 6 = 2$ and $y = -4 - 6 = -10$. Since a single input ($x=0$) leads to two different outputs ($y=2$ and $y=-10$), this equation does not define $y$ as a function of $x$. Graphically, this equation represents a parabola opening to the right, and as we discussed, a parabola that opens horizontally will fail the Vertical Line Test.

Equation 2: $6 x+|y|=0$

Our second equation is $6x + |y| = 0$. This equation involves an absolute value, which often requires careful consideration. Let's try to solve for $y$ again.

First, isolate the absolute value term by subtracting $6x$ from both sides:

$|y| = -6x$

Now, we need to think about the properties of absolute value. The absolute value of a number, $|y|$, is always non-negative (i.e., $|y| \ge 0$). This means that the expression on the right side, $-6x$, must also be non-negative.

$-6x \ge 0$

To satisfy this inequality, $x$ must be less than or equal to zero ($x \le 0$). If $x$ were a positive number, then $-6x$ would be negative, and we would have a situation where $|y|$ equals a negative number, which is impossible.

So, for any $x > 0$, there is no solution for $y$. This means the domain of this relation is restricted to $x \le 0$.

Now, let's consider the case where $x \le 0$. If $|y| = -6x$, what are the possible values for $y$? By the definition of absolute value, if $|y| = k$ (where $k \ge 0$), then $y = k$ or $y = -k$. In our case, $k = -6x$. So, we have:

$y = -6x$ or $y = -(-6x)$

$y = -6x$ or $y = 6x$

Let's test this with an example. Suppose we choose $x = -1$. Since $-1 \le 0$, this is a valid input. Plugging it into $|y| = -6x$, we get $|y| = -6(-1) = 6$. This means $y$ could be 6 or $y$ could be -6. Both of these values satisfy $|y|=6$.

Since a single input value ($x=-1$) leads to two possible output values ($y=6$ and $y=-6$), the equation $6x + |y| = 0$ does not define $y$ as a function of $x$. This relationship is only defined for $x \le 0$, but within that domain, it still fails the function test.

Equation 3: $y=4|x|-1$

Let's examine the third equation: $y = 4|x| - 1$. This equation already has $y$ isolated on one side, which is a good start. Now we just need to consider if, for any given $x$, there is only one value of $y$.

The key here is the absolute value of $x$, denoted as $|x|$. The absolute value function, by its very definition, always returns a single, non-negative value for any given input $x$. For instance, $|5| = 5$ and $|-5| = 5$. In both cases, the input $x$ yields a single output for $|x|$.

Since $|x|$ always produces a unique value for any $x$, multiplying it by 4 ($4|x|$) will also produce a unique value. Subtracting 1 ($4|x|-1$) will then result in a single, unique value for $y$.

Let's test this with a couple of values:

• If $x = 3$, then $y = 4|3| - 1 = 4(3) - 1 = 12 - 1 = 11$. We get one $y$ value.
• If $x = -3$, then $y = 4|-3| - 1 = 4(3) - 1 = 12 - 1 = 11$. We get the same single $y$ value.
• If $x = 0$, then $y = 4|0| - 1 = 4(0) - 1 = 0 - 1 = -1$. Again, a single $y$ value.

Because for every input $x$, there is exactly one output $y$, the equation $y = 4|x| - 1$ defines $y$ as a function of $x$. This type of function is related to the absolute value function, often resulting in a V-shaped graph.

Equation 4: x=rac{y^2}{6}

Finally, let's consider the fourth equation: x = rac{y^2}{6}. Our task is to determine if this equation defines $y$ as a function of $x$. To do this, we need to see if we can solve for $y$ and if each $x$ value produces only one $y$ value.

First, let's isolate the term involving $y$. We can do this by multiplying both sides of the equation by 6:

$6x = y^2$

Now, to solve for $y$, we need to take the square root of both sides. Just like in the first equation, taking the square root introduces the possibility of two solutions:

$\pm\sqrt{6x} = y$

So, we can write this as:

$y = \pm\sqrt{6x}$

For this equation to have real solutions for $y$, the expression under the square root, $6x$, must be non-negative. This means $6x \ge 0$, which implies $x \ge 0$. So, the domain for this relation is $x \ge 0$.

Now, let's consider if for a given $x$ (where $x \ge 0$), we get a unique $y$. If we choose any positive value for $x$, say $x=6$, then:

$y = \pm\sqrt{6 \times 6} = \pm\sqrt{36} = \pm 6$

This means that for $x=6$, we have two possible values for $y$: $y=6$ and $y=-6$. Since a single input ($x=6$) results in two different outputs, this equation does not define $y$ as a function of $x$. Graphically, this equation represents a parabola that opens to the right, and like the first equation, it fails the Vertical Line Test.

Conclusion: Identifying Functions

We've worked through four different equations, applying the principles of function definition. Remember, the core idea is that for an equation to define $y$ as a function of $x$, every valid input $x$ must correspond to exactly one output $y$.

• Equation 1: $(y+4)^2-36=x$: Did not define $y$ as a function of $x$ because solving for $y$ resulted in $y = -4 \pm\sqrt{x+36}$, yielding two $y$ values for many $x$ values.
• Equation 2: $6 x+|y|=0$: Did not define $y$ as a function of $x$. When $x < 0$, $|y| = -6x$ led to two possible values for $y$ ($y=-6x$ and $y=6x$).
• Equation 3: $y=4|x|-1$: Did define $y$ as a function of $x$. The absolute value $|x|$ always produces a single value for any given $x$, ensuring a unique $y$ output.
• Equation 4: x=rac{y^2}{6}: Did not define $y$ as a function of $x$. Solving for $y$ gave $y = \pm\sqrt{6x}$, resulting in two $y$ values for positive $x$ values.

Mastering the concept of functions is essential in mathematics. Understanding how to identify them from equations, graphs, and other representations will serve you well in further studies. For more in-depth exploration of functions and their properties, you might find the resources at Khan Academy very helpful.