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Choose the inverse of [tex]\( y = x^2 - 10x \)[/tex].

A. [tex]\( y = \pm \sqrt{x - 25} - 5 \)[/tex]
B. [tex]\( y = \pm \sqrt{x - 25} + 5 \)[/tex]
C. [tex]\( y = \pm \sqrt{x + 25} - 5 \)[/tex]
D. [tex]\( y = \pm \sqrt{x + 25} + 5 \)[/tex]


Sagot :

To find the inverse of the function [tex]\( y = x^2 - 10x \)[/tex], we need to follow several steps. Let's go through these steps one by one:

1. Rewrite the function: The given function is [tex]\( y = x^2 - 10x \)[/tex].

2. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]: To do this, we'll first rewrite the function in a way that isolates [tex]\( x \)[/tex]. Notice that we have a quadratic equation. We can rewrite it as:

[tex]\[ y = x^2 - 10x \][/tex]

We need to complete the square:

[tex]\[ y = x^2 - 10x \][/tex]

[tex]\[ y = (x^2 - 10x + 25) - 25 \][/tex]

[tex]\[ y = (x - 5)^2 - 25 \][/tex]

Now we have:

[tex]\[ y + 25 = (x - 5)^2 \][/tex]

3. Take the square root of both sides to solve for [tex]\( x \)[/tex]:

[tex]\[ \sqrt{y + 25} = |x - 5| \][/tex]

This gives us two possible solutions because the absolute value function results in both positive and negative solutions:

[tex]\[ x - 5 = \sqrt{y + 25} \quad \text{or} \quad x - 5 = -\sqrt{y + 25} \][/tex]

Solving for [tex]\( x \)[/tex] in both cases:

[tex]\[ x = 5 + \sqrt{y + 25} \quad \text{or} \quad x = 5 - \sqrt{y + 25} \][/tex]

4. Express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]: The inverse function [tex]\( y \)[/tex] should be expressed in terms of [tex]\( x \)[/tex]. We can now rewrite these as the inverse functions:

[tex]\[ y = \sqrt{x - 25} - 5 \quad \text{or} \quad y = -\sqrt{x - 25} - 5 \][/tex]

5. Match with given options:

- Option 1: [tex]\( y = \pm \sqrt{x - 25} - 5 \)[/tex]
- Option 2: [tex]\( y = \pm \sqrt{x - 25} + 5 \)[/tex]
- Option 3: [tex]\( y = \pm \sqrt{x + 25} - 5 \)[/tex]
- Option 4: [tex]\( y = \pm \sqrt{x + 25} + 5 \)[/tex]

Notice that our derived inverse functions match the forms found in Option 1.

Thus, the correct inverse of [tex]\( y = x^2 - 10x \)[/tex] is given by:

[tex]\[ y = \pm \sqrt{x - 25} - 5 \][/tex]