Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Our Q&A platform offers a seamless experience for finding reliable answers from experts in various disciplines. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
To find the inverse of the function [tex]\( y = x^2 - 10x \)[/tex], follow these steps:
1. Rewrite the original function:
[tex]\[ y = x^2 - 10x \][/tex]
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ x = y^2 - 10y \][/tex]
3. Reformat the equation to zero on one side:
[tex]\[ y^2 - 10y - x = 0 \][/tex]
4. Consider solving for [tex]\( y \)[/tex]. This is a quadratic equation in terms of [tex]\( y \)[/tex]:
[tex]\[ y^2 - 10y - x = 0 \][/tex]
5. Use the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -10 \)[/tex], and [tex]\( c = -x \)[/tex]:
[tex]\[ y = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(-x)}}{2(1)} \][/tex]
[tex]\[ y = \frac{10 \pm \sqrt{100 + 4x}}{2} \][/tex]
[tex]\[ y = \frac{10 \pm \sqrt{4(x + 25)}}{2} \][/tex]
[tex]\[ y = \frac{10 \pm 2\sqrt{x + 25}}{2} \][/tex]
[tex]\[ y = 5 \pm \sqrt{x + 25} \][/tex]
6. Thus, there are two possible solutions for the inverse function:
[tex]\[ y = 5 + \sqrt{x + 25} \][/tex]
[tex]\[ y = 5 - \sqrt{x + 25} \][/tex]
Upon simplifying, the inverse functions are:
[tex]\[ y = \sqrt{x + 25} + 5 \][/tex]
and
[tex]\[ y = \sqrt{x + 25} - 5 \][/tex]
Given these two forms, we compare the given choices:
- [tex]\( y = \pm \sqrt{x-25}-5 \)[/tex]
- [tex]\( y = \pm \sqrt{x-25}+5 \)[/tex]
- [tex]\( y = \pm \sqrt{x+25}-5 \)[/tex]
- [tex]\( y = \pm \sqrt{x+25}+5 \)[/tex]
The correct forms derived are:
[tex]\[ y = \pm \sqrt{x + 25} - 5 \][/tex]
and
[tex]\[ y = \pm \sqrt{x + 25} + 5 \][/tex]
So, the correct answer is:
[tex]\[ y = \pm \sqrt{x + 25} - 5 \][/tex]
1. Rewrite the original function:
[tex]\[ y = x^2 - 10x \][/tex]
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ x = y^2 - 10y \][/tex]
3. Reformat the equation to zero on one side:
[tex]\[ y^2 - 10y - x = 0 \][/tex]
4. Consider solving for [tex]\( y \)[/tex]. This is a quadratic equation in terms of [tex]\( y \)[/tex]:
[tex]\[ y^2 - 10y - x = 0 \][/tex]
5. Use the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -10 \)[/tex], and [tex]\( c = -x \)[/tex]:
[tex]\[ y = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(-x)}}{2(1)} \][/tex]
[tex]\[ y = \frac{10 \pm \sqrt{100 + 4x}}{2} \][/tex]
[tex]\[ y = \frac{10 \pm \sqrt{4(x + 25)}}{2} \][/tex]
[tex]\[ y = \frac{10 \pm 2\sqrt{x + 25}}{2} \][/tex]
[tex]\[ y = 5 \pm \sqrt{x + 25} \][/tex]
6. Thus, there are two possible solutions for the inverse function:
[tex]\[ y = 5 + \sqrt{x + 25} \][/tex]
[tex]\[ y = 5 - \sqrt{x + 25} \][/tex]
Upon simplifying, the inverse functions are:
[tex]\[ y = \sqrt{x + 25} + 5 \][/tex]
and
[tex]\[ y = \sqrt{x + 25} - 5 \][/tex]
Given these two forms, we compare the given choices:
- [tex]\( y = \pm \sqrt{x-25}-5 \)[/tex]
- [tex]\( y = \pm \sqrt{x-25}+5 \)[/tex]
- [tex]\( y = \pm \sqrt{x+25}-5 \)[/tex]
- [tex]\( y = \pm \sqrt{x+25}+5 \)[/tex]
The correct forms derived are:
[tex]\[ y = \pm \sqrt{x + 25} - 5 \][/tex]
and
[tex]\[ y = \pm \sqrt{x + 25} + 5 \][/tex]
So, the correct answer is:
[tex]\[ y = \pm \sqrt{x + 25} - 5 \][/tex]
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.