How G(x) = F(x) + 5 Shifts The Graph Of F(x)

When we talk about comparing the graph of $g(x)$ with the graph of $f(x)$, especially when $g(x)$ is defined as $f(x) + 5$, we're delving into the fascinating world of function transformations. These transformations allow us to understand how changes to a function's equation affect its visual representation on a graph. In this specific case, the relationship $g(x) = f(x) + 5$ tells us something very precise about how the graph of $g(x)$ relates to the graph of $f(x)$. Let's break down why option A, 'The graph of $g(x)$ is the graph of $f(x)$ shifted 5 units up,' is the best and only correct statement that compares these two graphs. We'll explore the underlying mathematical principles that govern these vertical shifts and why other potential transformations, like horizontal shifts, don't apply here. Understanding these concepts is crucial for anyone studying algebra and calculus, as it provides a powerful toolkit for analyzing and predicting the behavior of functions. By mastering these transformations, you gain the ability to visualize complex functions more easily and to understand the impact of simple additions or subtractions on their graphical output. It’s like having a secret decoder ring for understanding mathematical graphs, allowing you to see the connections between different functions with clarity and confidence.

To truly grasp why the graph shifts vertically, let's consider what happens to the output values (the $y$-values) of the function when we move from $f(x)$ to $g(x)$. For any given input value, say $x_0$, the output of $f(x)$ is $f(x_0)$. Now, according to the definition of $g(x)$, the output for the same input $x_0$ is $g(x_0) = f(x_0) + 5$. This means that for every single input value, the corresponding output of $g(x)$ is exactly 5 units greater than the output of $f(x)$. If $f(x_0)$ produced a $y$-value of, let's say, 10, then $g(x_0)$ will produce a $y$-value of $10 + 5 = 15$. If $f(x_0)$ produced a $y$-value of -3, then $g(x_0)$ will produce a $y$-value of $-3 + 5 = 2$. This consistent increase of 5 in the $y$-value, regardless of the input $x$, directly translates to a vertical movement of the entire graph. Imagine plotting points $(x, f(x))$ for the function $f(x)$. Now, for the same $x$-values, you are plotting points $(x, f(x) + 5)$, which is $(x, g(x))$. Each point on the graph of $f(x)$ is being lifted exactly 5 units upwards to create the corresponding point on the graph of $g(x)$. This is precisely what a vertical shift upwards by 5 units means in graphical terms. The shape and orientation of the graph remain identical; only its position along the $y$-axis changes. This principle is fundamental and applies to all types of functions, whether they are linear, quadratic, trigonometric, or exponential. The addition of a constant to the function's output universally results in a vertical translation.

Now, let's quickly address why the other options are incorrect to solidify our understanding. Option B suggests that 'The graph of $g(x)$ is the graph of $f(x)$ shifted 5 units to the left.' A horizontal shift to the left by 5 units would be represented by a function like $h(x) = f(x + 5)$. Notice the change occurs inside the function's argument, not outside. When we replace $x$ with $(x+5)$ in $f(x)$, we are essentially asking the function to evaluate itself at a point 5 units to the right of the current $x$ to achieve the same output value. This results in the entire graph moving 5 units to the left. Since our transformation is $g(x) = f(x) + 5$, where the '+ 5' is outside the $f( ule{0.5cm}{0.4pt})$ operation, it affects the output ($y$-value), not the input ($x$-value). Therefore, a horizontal shift is not what's happening here. Similarly, other transformations like horizontal stretches or compressions (e.g., $g(x) = f(5x)$ or $g(x) = f(x/5)$) or vertical stretches and compressions (e.g., $g(x) = 5f(x)$ or g(x) = rac{1}{5}f(x)) would involve multiplication or division factors applied to the input or output, respectively. The simple addition of a constant to the function's output is the defining characteristic of a vertical translation. This makes the statement 'The graph of $g(x)$ is the graph of $f(x)$ shifted 5 units up' the most accurate and descriptive comparison.

Consider a concrete example to illustrate this principle. Let $f(x) = x^2$. This is a standard parabola opening upwards with its vertex at the origin $(0,0)$. Now, let $g(x) = f(x) + 5$. Substituting the definition of $f(x)$, we get $g(x) = x^2 + 5$. Let's look at some key points:

• For $f(x) = x^2$:

  • At $x=0$, $f(0) = 0^2 = 0$. Point: $(0, 0)$.
  • At $x=1$, $f(1) = 1^2 = 1$. Point: $(1, 1)$.
  • At $x=-1$, $f(-1) = (-1)^2 = 1$. Point: $(-1, 1)$.
  • At $x=2$, $f(2) = 2^2 = 4$. Point: $(2, 4)$.

• For $g(x) = x^2 + 5$:

  • At $x=0$, $g(0) = 0^2 + 5 = 5$. Point: $(0, 5)$.
  • At $x=1$, $g(1) = 1^2 + 5 = 6$. Point: $(1, 6)$.
  • At $x=-1$, $g(-1) = (-1)^2 + 5 = 6$. Point: $(-1, 6)$.
  • At $x=2$, $g(2) = 2^2 + 5 = 9$. Point: $(2, 9)$.

Comparing the points, we see that for every $x$-value, the $y$-value of $g(x)$ is exactly 5 more than the $y$-value of $f(x)$. The vertex of $f(x)$ is at $(0,0)$, while the vertex of $g(x)$ is at $(0,5)$. All the points on the graph of $g(x)$ are 5 units higher than their corresponding points on the graph of $f(x)$. This confirms that the graph of $g(x) = x^2 + 5$ is indeed the graph of $f(x) = x^2$ shifted 5 units upward. This visual and numerical evidence strongly supports the conclusion. The consistency across different input values reinforces the understanding that the '+5' directly impacts the vertical position of the graph.

This concept of vertical translation is fundamental in understanding how functions behave and how their graphs are altered. When a constant is added outside the function's operation, it universally results in a vertical shift. If the constant is positive, the shift is upwards; if it's negative, the shift is downwards. This is because the constant directly modifies the output ($y$-value) of the function for every input ($x$-value). It's a straightforward additive change to the function's height. Conversely, if the constant were added or subtracted inside the function's argument (e.g., $f(x+c)$ or $f(x-c)$), it would cause a horizontal shift. A positive $c$ in $f(x+c)$ shifts the graph to the left, while a positive $c$ in $f(x-c)$ shifts the graph to the right. This distinction between operations inside and outside the function's parentheses is key to correctly identifying transformations. In our specific case, $g(x) = f(x) + 5$, the '+ 5' is clearly outside the $f( ule{0.5cm}{0.4pt})$ operation, acting directly on the output. This makes it a pure vertical translation. No other transformation is implied by this simple additive relationship. The shape of the parabola, for instance, remains exactly the same; only its position on the $y$-axis is adjusted. This allows us to quickly sketch or understand transformed functions without needing to calculate numerous points, provided we understand the basic rules of transformation. It's a powerful shortcut in mathematical analysis and visualization.

In summary, when you see a function defined as $g(x) = f(x) + c$, where $c$ is a constant, you should immediately recognize this as a vertical translation of the graph of $f(x)$. If $c > 0$, the graph shifts upward by $c$ units. If $c < 0$, the graph shifts downward by $|c|$ units. In the specific instance of $g(x) = f(x) + 5$, the constant is positive ($+5$), indicating an upward shift. Therefore, the statement that best compares the graph of $g(x)$ with the graph of $f(x)$ is that the graph of $g(x)$ is the graph of $f(x)$ shifted 5 units up. This is a fundamental rule in function transformations that is essential for understanding more complex graphical manipulations. This principle extends to all types of functions and is a cornerstone of graphical analysis in mathematics. Understanding these transformations is key to not just passing exams but to truly comprehending how mathematical functions operate and how their visual representations convey meaning.

For more in-depth information on function transformations, you can explore resources from Khan Academy which offers comprehensive explanations and exercises on this topic.