The problem requires finding the minimum value of the quadratic function $y = x^2 - 7$ over the range $-2 < x < 2$, and the answer should be rounded to three decimal places.

Step 1: Understanding the Function

The function is $y = x^2 - 7$, a quadratic equation with:

• $x^2$ being the main term that determines the shape of the curve.
• The constant $-7$ shifts the parabola downward.

Step 2: Characteristics of the Parabola

• $y = x^2 - 7$ is a parabola opening upwards.
• Its vertex is at $x = 0$, which is the lowest point of the parabola.

Step 3: Range of Interest

We are given the range $-2 < x < 2$. Since the vertex $x = 0$ lies within this range, the minimum value of the function will occur at the vertex.

Step 4: Calculate the Minimum Value

Substitute $x = 0$ into the equation: $y = 0^2 - 7 = -7$

Final Answer:

The minimum value of $y = x^2 - 7$ over the range $-2 < x < 2$ is $-7.000$ (to three decimal places).

Let me know if you need a detailed explanation or have further questions! 😊

Here are 5 related questions to deepen your understanding:

1. How do you determine if a parabola opens upwards or downwards?
2. What is the significance of the vertex in a quadratic equation?
3. Can the minimum value of a parabola occur outside the vertex in certain intervals?
4. How would the range affect the minimum value if it excluded $x = 0$?
5. What are the steps to find the maximum value of a function?

Tip: Always graph your function or analyze its vertex and direction to ensure accuracy when finding extremum values.