To find the inverse of the function:

$y = -4x^2 + 3(2x + 1) + 2$

Step-by-step solution:

1. Simplify the given function:
   Expand the expression on the right-hand side.

$y = -4x^2 + 3(2x + 1) + 2$ $y = -4x^2 + 6x + 3 + 2$ $y = -4x^2 + 6x + 5$

Now, we have:

$y = -4x^2 + 6x + 5$

2. Switch $x$ and $y$ to find the inverse function: We will solve for $x$ in terms of $y$ by interchanging the variables.

$x = -4y^2 + 6y + 5$

Now, solve this quadratic equation for $y$ in terms of $x$.

$-4y^2 + 6y + (5 - x) = 0$

This is a quadratic equation in $y$, and we can solve it using the quadratic formula:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where:

• $a = -4$
• $b = 6$
• $c = 5 - x$

Substitute the values into the quadratic formula:

$y = \frac{-6 \pm \sqrt{6^2 - 4(-4)(5 - x)}}{2(-4)}$ $y = \frac{-6 \pm \sqrt{36 + 16(5 - x)}}{-8}$ $y = \frac{-6 \pm \sqrt{36 + 80 - 16x}}{-8}$ $y = \frac{-6 \pm \sqrt{116 - 16x}}{-8}$

Thus, the inverse function is:

$y = \frac{-6 \pm \sqrt{116 - 16x}}{-8}$

Since there are two possible values for the inverse due to the quadratic nature of the original function, it may not have a unique inverse. For clarity, we would often restrict the domain to a section of the original function to ensure it is one-to-one.

Would you like more details or have any questions?

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Here are 5 related questions:

1. How do you determine the domain and range of the inverse function?
2. What is the significance of restricting the domain in quadratic functions?
3. How can the quadratic formula be derived from completing the square?
4. How do you graph both the original and inverse functions?
5. What are real-world applications of inverse functions?

Tip: When solving for inverses, remember that not all functions are invertible unless they are one-to-one. Always check for restrictions on the domain.