https://file.aiquickdraw.com/mathbot/file/1727204022029e5zd42hn.jpg

Let's analyze the function \( h(x) \) from the image:

\[
h(x) =
\begin{cases}
2 + x, & x \leq 4 \\
10x - x^2 - 17, & x > 4
\end{cases}
\]

### a) Sketch the graph of \( y = h(x) \)

For \( x \leq 4 \), the function is linear: \( h(x) = 2 + x \). This is a straight line with a slope of 1 and y-intercept of 2.

For \( x > 4 \), the function is quadratic: \( h(x) = 10x - x^2 - 17 \). This is a parabola that opens downward due to the negative coefficient of \( x^2 \).

Key points to consider when sketching:

- At \( x = 4 \), evaluate both parts of the function:
  - \( h(4) = 2 + 4 = 6 \) for the first part.
  - For the second part, also evaluate \( h(4) = 10(4) - 4^2 - 17 = 40 - 16 - 17 = 7 \).
  - So, the function has a **jump discontinuity** at \( x = 4 \) from 6 to 7.

- The graph includes:
  - A line for \( x \leq 4 \),
  - A downward-opening parabola for \( x > 4 \),
  - A discontinuity at \( x = 4 \).

### b) State the value of \( h(4) \).

From the first part of the function (for \( x \leq 4 \)), \( h(4) = 6 \).

### c) State the range of \( h(x) \).

For the first part, \( h(x) = 2 + x \) is linear and increases without bound for \( x \leq 4 \), so the maximum value is \( h(4) = 6 \).

For the second part, the parabola \( 10x - x^2 - 17 \) opens downward, so it has a maximum value at its vertex.

- The vertex occurs at \( x = -\frac{b}{2a} = \frac{10}{2(1)} = 5 \),
- Substituting \( x = 5 \) into the function gives \( h(5) = 10(5) - 5^2 - 17 = 50 - 25 - 17 = 8 \).

So, the range is \( (-\infty, 8] \), since the function reaches a maximum of 8 at \( x = 5 \) and decreases as \( x \to \infty \).

### d) Find the exact values of \( x \) for which \( h(x) = 5 \).

For \( h(x) = 5 \), we need to solve for both parts of the piecewise function:

1. First part: \( 2 + x = 5 \)
   \[
   x = 5 - 2 = 3.
   \]
   So, \( x = 3 \) is one solution.

2. Second part: \( 10x - x^2 - 17 = 5 \)
   \[
   10x - x^2 - 17 = 5 \quad \Rightarrow \quad -x^2 + 10x - 22 = 0 \quad \Rightarrow \quad x^2 - 10x + 22 = 0.
   \]
   Solving this quadratic using the quadratic formula:
   \[
   x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(22)}}{2(1)} = \frac{10 \pm \sqrt{100 - 88}}{2} = \frac{10 \pm \sqrt{12}}{2} = \frac{10 \pm 2\sqrt{3}}{2}.
   \]
   Simplifying, we get:
   \[
   x = 5 \pm \sqrt{3}.
   \]

Thus, the exact values of \( x \) for which \( h(x) = 5 \) are \( x = 3 \), \( x = 5 + \sqrt{3} \), and \( x = 5 - \sqrt{3} \).

---

Would you like any further explanation on these parts? 

Here are some follow-up questions to deepen your understanding:

1. What is the domain of \( h(x) \)?
2. How does the behavior of the graph change at \( x = 4 \)?
3. Can you determine the vertex of the parabola section of \( h(x) \)?
4. What type of discontinuity occurs at \( x = 4 \)?
5. How would the range change if the quadratic part of the function were adjusted?

**Tip**: When dealing with piecewise functions, always check for discontinuities at the transition points to determine if the function is continuous or not!

The function h(x) is defined by two parts: h(x) = 2 + x for x ≤ 4, and h(x) = 10x - x^2 - 17 for x > 4. The task is to sketch the graph, find h(4), determine the range, and solve for when h(x) = 5.

Learn how to analyze and solve a piecewise function that combines a linear and quadratic function. This problem involves sketching the graph, identifying the range, and solving for specific function values, making it ideal for Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation