Piecewise Functions: Finding The Y-Intercept

When we talk about piecewise functions, we're dealing with functions that are defined by different formulas over different intervals of their domain. Think of it like a choose-your-own-adventure book, but for math! Each "piece" of the function is like a different path you can take, depending on the value of x you're currently looking at. This makes them super versatile for modeling real-world situations where behavior changes. For instance, consider the cost of postage, which might have different rates for different weight increments, or the temperature in a room controlled by a thermostat that switches on and off at certain temperatures. Understanding how to work with these functions is key to unlocking their power. One of the most fundamental things we often want to find with any function is its y-intercept. This is the point where the graph of the function crosses the y-axis. Mathematically, this occurs when x is equal to 0. So, to find the y-intercept of a piecewise function, we need to determine which "piece" of the function is active when x = 0.

Let's dive into the specific function you've presented:

$$f(x)=\left\{\begin{array}{rcr} 7 x-3 & \text { if } & x<2 \ x^2+5 & \text { if } & 2 \leq x<11 \ 6 & \text { if } & x \geq 11 \end{array}\right.$$

To find the y-intercept, we need to evaluate the function at x = 0. The crucial step here is to identify which of the conditions given in the function definition is satisfied when x = 0. Let's examine each piece:

• Piece 1: $7x - 3$ is defined for $x < 2$. Does x = 0 satisfy this condition? Yes, because 0 is indeed less than 2. This means that this piece of the function is the one we should use to calculate the y-intercept.
• Piece 2: $x^2 + 5$ is defined for $2 \leq x < 11$. Does x = 0 satisfy this condition? No, because 0 is not greater than or equal to 2.
• Piece 3: $6$ is defined for $x \geq 11$. Does x = 0 satisfy this condition? No, because 0 is not greater than or equal to 11.

Since x = 0 falls within the interval $x < 2$, we use the first piece of the function, $f(x) = 7x - 3$, to find the y-intercept. Plugging in x = 0 into this expression, we get:

$f(0) = 7(0) - 3 = 0 - 3 = -3$

Therefore, the y-intercept of this piecewise function is -3. The coordinates of the y-intercept are (0, -3). It's important to remember that a piecewise function can only have a y-intercept if x = 0 falls into one of the defined intervals. In this case, it clearly does, and it falls into only one interval, ensuring a unique y-intercept. If x = 0 had not fallen into any of the intervals, the function would not have a y-intercept. For example, if the first condition was $x < -1$, then $x=0$ would not satisfy any of the conditions.

The key takeaway here is to always check the conditions for each piece of the function against the x-value you are interested in. For the y-intercept, that x-value is always 0. By systematically checking each condition, you can confidently determine which formula to apply and avoid errors. This methodical approach ensures that you are using the correct part of the function's definition, which is the cornerstone of working with piecewise functions accurately. The visual representation of a piecewise function can also be helpful. Imagine drawing the graph of $y = 7x - 3$. This is a straight line. Then, you would only keep the part of this line where x values are less than 2. Next, you'd graph $y = x^2 + 5$. This is a parabola. You'd keep the portion of the parabola where x is between 2 (inclusive) and 11 (exclusive). Finally, you'd graph $y = 6$. This is a horizontal line, and you'd keep only the part where x is 11 or greater. The y-intercept is where the entire combined graph crosses the y-axis, which happens at x = 0. Since our combined graph starts with the line segment from $y = 7x - 3$ at $x=0$, that's where we find our intercept.

Understanding the domain for each piece is absolutely critical. Let's re-emphasize this. The conditions $x < 2$, $2 \leq x < 11$, and $x \geq 11$ partition the entire number line. Every possible real number x falls into exactly one of these three categories. This is important because it guarantees that for any given x, there's a unique rule that applies. When we're looking for the y-intercept, we're specifically interested in the behavior of the function at $x=0$. We ask ourselves: "Where does $x=0$ fit in?" Looking at the conditions:

1. Is $0 < 2$? Yes. This piece applies.
2. Is $2 \leq 0 < 11$? No. This piece does not apply.
3. Is $0 \geq 11$? No. This piece does not apply.

Because $x=0$ satisfies the condition for the first piece ($x < 2$), we use the formula associated with that piece, which is $f(x) = 7x - 3$. Substituting $x=0$ into this formula gives us $f(0) = 7(0) - 3 = -3$. The y-intercept is the point $(0, -3)$.

It's worth noting that sometimes a piecewise function might be defined in a way that $x=0$ doesn't fall into any of the specified intervals. For instance, if the function was defined as:

$$g(x)=\left\{\begin{array}{rcr} 7 x-3 & \text { if } & x< -5 \ x^2+5 & \text { if } & -5 \leq x<0 \ 6 & \text { if } & x \geq 0 \end{array}\right.$$

In this modified g(x) function, $x=0$ satisfies the condition $x \geq 0$, so we would use the third piece, $f(x)=6$. Then $f(0)=6$, and the y-intercept would be $(0,6)$. However, if the function was:

$$h(x)=\left\{\begin{array}{rcr} 7 x-3 & \text { if } & x< -5 \ x^2+5 & \text { if } & -5 \leq x<-2 \ 6 & \text { if } & x \geq -2 \end{array}\right.$$

Here, $x=0$ satisfies $x \geq -2$, so we would use the third piece, $f(x)=6$. Then $f(0)=6$, and the y-intercept would be $(0,6)$. If $x=0$ did not fall into any interval, then the function would have no y-intercept. The question asks which "piece(s)" should be used. For the given function $f(x)$, the answer is straightforward: only the first piece, defined by $7x-3$ for $x < 2$, is used because $x=0$ satisfies the condition $x < 2$. The other pieces are irrelevant for finding the y-intercept in this specific case. Always remember to test $x=0$ against the interval conditions first!

Identifying the Correct Interval

The first and most crucial step in finding the y-intercept of a piecewise function is to correctly identify which interval contains $x=0$. This is the gatekeeper to using the right formula. For the function provided:

$$f(x)=\left\{\begin{array}{rcr} 7 x-3 & \text { if } & x<2 \ x^2+5 & \text { if } & 2 \leq x<11 \ 6 & \text { if } & x \geq 11 \end{array}\right.$$

We test $x=0$ against each condition:

• Condition 1: $x < 2$. Is $0 < 2$? Yes. This condition is met.
• Condition 2: $2 \leq x < 11$. Is $2 \leq 0 < 11$? No. 0 is not greater than or equal to 2.
• Condition 3: $x \geq 11$. Is $0 \geq 11$? No. 0 is not greater than or equal to 11.

Since $x=0$ satisfies only the first condition ($x < 2$), we know that the first piece of the function, $f(x) = 7x - 3$, is the one that governs the behavior of the function at $x=0$. It is the only piece we need to consider for calculating the y-intercept.

Calculating the Y-Intercept Value

Once we've identified the correct piece, the calculation is usually straightforward. As established, for the given function, the first piece ($f(x) = 7x - 3$) is used when $x=0$. To find the y-intercept, we substitute $x=0$ into this expression:

$f(0) = 7(0) - 3$

$f(0) = 0 - 3$

$f(0) = -3$

So, the y-intercept is the point where the function crosses the y-axis, which has coordinates $(0, -3)$. The question asks which "piece(s)" should be used. Based on our analysis, the answer is unequivocally the first piece. It is vital to understand that a piecewise function, by its nature, assigns a single output for any given input that falls within its domain. Since $x=0$ falls exclusively into the interval $x < 2$, only the formula corresponding to that interval ($7x-3$) is relevant for determining $f(0)$.

It's essential to be precise. The question asks for the