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Let's begin by solving the problems 1-47 and 1-48 based on the image you provided.

### 1-47:
We are given \( f(x) = \frac{1}{x} \) and asked to compute the following:

**a. \( f\left(\frac{1}{2}\right) \):**

\[
f\left(\frac{1}{2}\right) = \frac{1}{\frac{1}{2}} = 2
\]

**b. \( f(1) \):**

\[
f(1) = \frac{1}{1} = 1
\]

**c. \( f(0.01) \):**

\[
f(0.01) = \frac{1}{0.01} = 100
\]

**d. \( f(0.007) \):**

\[
f(0.007) = \frac{1}{0.007} \approx 142.857
\]

### 1-48:
We are asked to solve the following quadratic equations:

**a. \( x^2 - 8x + 15 = 0 \):**

To solve this, we use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
For the equation \( x^2 - 8x + 15 = 0 \), we have \( a = 1 \), \( b = -8 \), and \( c = 15 \).

First, calculate the discriminant:
\[
\Delta = (-8)^2 - 4(1)(15) = 64 - 60 = 4
\]

Now, use the quadratic formula:
\[
x = \frac{-(-8) \pm \sqrt{4}}{2(1)} = \frac{8 \pm 2}{2}
\]

Thus, the two solutions are:
\[
x_1 = \frac{8 + 2}{2} = 5 \quad \text{and} \quad x_2 = \frac{8 - 2}{2} = 3
\]

So, the solutions are \( x = 5 \) and \( x = 3 \).

**b. \( 2x^2 - 5x - 6 = 0 \):**

Using the quadratic formula again:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
For the equation \( 2x^2 - 5x - 6 = 0 \), we have \( a = 2 \), \( b = -5 \), and \( c = -6 \).

First, calculate the discriminant:
\[
\Delta = (-5)^2 - 4(2)(-6) = 25 + 48 = 73
\]

Now, apply the quadratic formula:
\[
x = \frac{-(-5) \pm \sqrt{73}}{2(2)} = \frac{5 \pm \sqrt{73}}{4}
\]

Thus, the solutions are:
\[
x_1 = \frac{5 + \sqrt{73}}{4} \quad \text{and} \quad x_2 = \frac{5 - \sqrt{73}}{4}
\]

These are the exact solutions, and you can approximate them numerically if needed.

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Would you like further details on these solutions or have any other questions?

Here are 5 related questions you might consider:
1. How is the quadratic formula derived?
2. What is the geometric interpretation of a quadratic equation's solutions?
3. Can you solve \( x^2 - 8x + 15 = 0 \) by factoring instead of using the quadratic formula?
4. What happens if the discriminant \( \Delta \) is negative in a quadratic equation?
5. Can quadratic equations always be solved using the quadratic formula, or are there exceptions?

**Tip:** Always check if a quadratic equation can be factored easily before using the quadratic formula, as factoring is often faster!

Solve the function f(x) = 1/x for different values, and quadratic equations like x^2 - 8x + 15 and 2x^2 - 5x - 6 = 0, using the quadratic formula. Suitable for students in Grades 7-10, this problem helps in understanding core algebra concepts.

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