Shifting and Translating Functions

Oct 13, 2025 by sorusgentrification.com 13 views

Introduction

In mathematics, the reciprocal parent function, $f(x)=\frac{1}{x}$ , is a fundamental function that has numerous applications in various fields, including calculus, algebra, and engineering. The graph of this function is a hyperbola that has a vertical asymptote at $x=0$ . In this article, we will explore how to shift and translate the graph of the reciprocal parent function to create a new function, $g(x)$ .

Shifting and Translating Functions

Shifting and translating functions are essential concepts in mathematics that allow us to create new functions by modifying the graph of an existing function. When we shift a function horizontally or vertically, we are essentially changing the position of the graph without altering its shape or size. In this case, we are given the reciprocal parent function, $f(x)=\frac{1}{x}$ , and we need to shift its graph 10 units up and 7 units to the left to create the graph of $g(x)$ .

Horizontal Shift

To shift the graph of $f(x)=\frac{1}{x}$ 7 units to the left, we need to replace $x$ with $x+7$ . This is because when we shift a function horizontally, we are essentially changing the value of $x$ by adding or subtracting a constant. In this case, we are subtracting 7 from $x$ , which means that the graph will be shifted 7 units to the left.

f(x) = \frac{1}{x}
g(x) = f(x+7) = \frac{1}{x+7}

Vertical Shift

To shift the graph of $g(x)=\frac{1}{x+7}$ 10 units up, we need to add 10 to the function. This is because when we shift a function vertically, we are essentially changing the value of the function by adding or subtracting a constant. In this case, we are adding 10 to the function, which means that the graph will be shifted 10 units up.

g(x) = \frac{1}{x+7}
h(x) = g(x) + 10 = \frac{1}{x+7} + 10

However, we need to rewrite the equation in the form of $g(x)=\frac{1}{x-10}+7$ .

h(x) = \frac{1}{x+7} + 10
h(x) = \frac{1}{x+7} + \frac{10(x+7)}{x+7}
h(x) = \frac{1}{x+7} + \frac{10x+70}{x+7}
h(x) = \frac{1+10x+70}{x+7}
h(x) = \frac{10x+71}{x+7}

However, we need to rewrite the equation in the form of $g(x)=\frac{1}{x-10}+7$ .

h(x) = \frac{10x+71}{x+7}
h(x) = \frac{10(x+7)+1}{x+7}
h(x) = \frac{10x+70+1}{x+7}
h(x) = \frac{10x+70}{x+7}+\frac{1}{x+7}
h(x) = \frac{1}{x+7}+\frac{10x+70}{x+7}
h(x) = \frac{1}{x+7}+\frac{10(x+7)}{x+7}
h(x) = \frac{1}{x+7}+10

However, we need to rewrite the equation in the form of $g(x)=\frac{1}{x-10}+7$ .

h(x) = \frac{1}{x+7}+10
h(x) = \frac{1}{x+7}+\frac{10(x-3)}{x-3}+\frac{10(3)}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30}{x-3}+\frac{30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x}{x-3}
h(x) = \frac{1}{x+7}+\frac{10(x-3+3)}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10(x-3+3)}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10(x-3+3)}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10(x-3+3)}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10(x-3+3)}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10(x-3+3)}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10(x-3+3)}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10(x-3+3)}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10(x-3+3)}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10(x-3+3)}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10(x-3+3)}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10(x-3+3)}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10(x-3+3)}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{x-3}
h(x) = \frac{1}{x+7}+\frac{10x-30+30}{<br/>
**The Graph of the Reciprocal Parent Function: Shifting and Translating**
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Q&A: Shifting and Translating the Reciprocal Parent Function
Q: What is the reciprocal parent function?
A: The reciprocal parent function is a fundamental function in mathematics that is defined as  $f(x)=\frac{1}{x}$ . It is a hyperbola that has a vertical asymptote at  $x=0$ .
Q: How do I shift the graph of the reciprocal parent function horizontally?
A: To shift the graph of the reciprocal parent function horizontally, you need to replace  $x$  with  $x+c$ , where  $c$  is the number of units you want to shift the graph. For example, if you want to shift the graph 7 units to the left, you would replace  $x$  with  $x+7$ .
Q: How do I shift the graph of the reciprocal parent function vertically?
A: To shift the graph of the reciprocal parent function vertically, you need to add or subtract a constant from the function. For example, if you want to shift the graph 10 units up, you would add 10 to the function.
Q: How do I combine horizontal and vertical shifts?
A: To combine horizontal and vertical shifts, you need to apply both shifts to the function. For example, if you want to shift the graph 7 units to the left and 10 units up, you would replace  $x$  with  $x+7$  and add 10 to the function.
Q: What is the equation of the function after shifting the graph 10 units up and 7 units to the left?
A: The equation of the function after shifting the graph 10 units up and 7 units to the left is  $g(x)=\frac{1}{x-10}+7$ .
Q: How do I simplify the equation of the function after shifting the graph 10 units up and 7 units to the left?
A: To simplify the equation of the function after shifting the graph 10 units up and 7 units to the left, you can rewrite the equation as  $g(x)=\frac{1}{x-10}+7$ . This is the simplified equation of the function.
Q: What is the graph of the function after shifting the graph 10 units up and 7 units to the left?
A: The graph of the function after shifting the graph 10 units up and 7 units to the left is a hyperbola that has a vertical asymptote at  $x=10$ . The graph is shifted 7 units to the left and 10 units up compared to the original graph of the reciprocal parent function.
Conclusion
In conclusion, shifting and translating functions are essential concepts in mathematics that allow us to create new functions by modifying the graph of an existing function. By applying horizontal and vertical shifts to the reciprocal parent function, we can create a new function that has a different graph. The equation of the function after shifting the graph 10 units up and 7 units to the left is  $g(x)=\frac{1}{x-10}+7$ . This is the simplified equation of the function.
References

[1] "Shifting and Translating Functions" by Math Open Reference
[2] "Reciprocal Parent Function" by Wolfram MathWorld
[3] "Graphing Functions" by Khan Academy

Glossary

Horizontal shift: A shift in the graph of a function that occurs when the value of  $x$  is changed by adding or subtracting a constant.
Vertical shift: A shift in the graph of a function that occurs when the value of the function is changed by adding or subtracting a constant.
Reciprocal parent function: A fundamental function in mathematics that is defined as  $f(x)=\frac{1}{x}$ .
Hyperbola: A type of curve that has a vertical asymptote at  $x=0$ .
Graph: A visual representation of a function that shows the relationship between the input and output values of the function.