To identify an appropriate quadratic model for the function $g(a) = \frac{1}{(a+3)^2} + \frac{1}{a+3} - 2$, we will follow a systematic approach:

Step 1: Simplify the function

Let’s simplify $g(a)$ by performing a substitution that will make it easier to work with:

Let: $u = a + 3$ This substitution rewrites the function $g(a)$ as: $g(a) = \frac{1}{u^2} + \frac{1}{u} - 2$

Now, the expression in terms of $u$ is: $g(u) = \frac{1}{u^2} + \frac{1}{u} - 2$

Step 2: Common denominator

Combine the first two terms by finding a common denominator: $g(u) = \frac{1 + u}{u^2} - 2$

Step 3: Expand and reorganize

Simplify this expression further: $g(u) = \frac{1 + u}{u^2} - 2 = \frac{1 + u - 2u^2}{u^2} = \frac{-2u^2 + u + 1}{u^2}$

Now the function is expressed as: $g(u) = \frac{-2u^2 + u + 1}{u^2}$

At this point, we can see that the function is not exactly a quadratic in the form of $au^2 + bu + c$, but we’ll analyze its behavior.

Step 4: Study maximums, minimums, and zeros

We will now analyze the properties:

1. Zeros: Set $g(u) = 0$ and solve for $u$. This would involve solving: $\frac{-2u^2 + u + 1}{u^2} = 0$ Which simplifies to: $-2u^2 + u + 1 = 0$ The discriminant of this quadratic equation is: $\Delta = 1^2 - 4(-2)(1) = 1 + 8 = 9$ Since the discriminant is positive, the equation has two real roots: $u = \frac{-1 \pm \sqrt{9}}{2(-2)} = \frac{-1 \pm 3}{-4}$ The two solutions are: $u_1 = \frac{-1 + 3}{-4} = \frac{2}{-4} = -\frac{1}{2}, \quad u_2 = \frac{-1 - 3}{-4} = \frac{-4}{-4} = 1$ Therefore, the zeros occur at $u = -\frac{1}{2}$ and $u = 1$.

2. Vertex and Extremes: For the quadratic $-2u^2 + u + 1$, the vertex can be found using the formula $u_{\text{vertex}} = \frac{-b}{2a}$, where $a = -2$ and $b = 1$: $u_{\text{vertex}} = \frac{-1}{2(-2)} = \frac{1}{4}$ The vertex occurs at $u = \frac{1}{4}$.

   Now, plug this $u$-value back into the function to find the corresponding $g(u)$: $g\left(\frac{1}{4}\right) = \frac{-2\left(\frac{1}{4}\right)^2 + \frac{1}{4} + 1}{\left(\frac{1}{4}\right)^2}$ Simplifying this gives the function value at the vertex, which corresponds to a minimum since the quadratic coefficient $a = -2$ indicates a downward-facing parabola.

3. Maximum values: Since this is a downward parabola, there is no global maximum, only a minimum at the vertex.

Step 5: Revert back to original variable $a$

Using the substitution $u = a + 3$, the zeros occur when: $a + 3 = -\frac{1}{2} \quad \text{or} \quad a + 3 = 1$ So: $a_1 = -\frac{7}{2}, \quad a_2 = -2$

The vertex occurs at $u = \frac{1}{4}$, so: $a_{\text{vertex}} = \frac{1}{4} - 3 = -\frac{11}{4}$

Conclusion:

• Zeros: $a = -\frac{7}{2}$ and $a = -2$.
• Minimum value: Occurs at $a = -\frac{11}{4}$.
• Maximum value: Does not exist.

Would you like the full details or have any other questions? Here are some related questions:

1. How do you derive the vertex of a quadratic function?
2. Can you explain how the discriminant determines the number of roots?
3. What is the significance of the quadratic's leading coefficient in finding maxima or minima?
4. How can we convert a complex rational expression into a quadratic form?
5. How do quadratic models help in analyzing real-world data?

Tip: For any quadratic function $ax^2 + bx + c$, the sign of $a$ (positive or negative) tells you whether the parabola opens upward or downward. This helps in determining if it has a minimum or maximum value!