Math Problem Statement
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.
Solution
Let's analyze the function from the image:
\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!Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Quadratic Functions
Algebra
Continuity
Range
Formulas
Piecewise function: h(x) = {2 + x for x ≤ 4, 10x - x^2 - 17 for x > 4}
Quadratic formula: x = [-b ± √(b^2 - 4ac)] / 2a
Theorems
Quadratic Formula
Continuity of Piecewise Functions
Suitable Grade Level
Grades 10-12
Related Recommendation
Identifying Piecewise Functions with Linear and Quadratic Components
Solving Piecewise Functions Based on Graph with Correct Equations for x Intervals
Defining a Piecewise Function from a Graph
Graphing and Finding the Range of a Piecewise Function
Piecewise Function Analysis with Absolute Value and Quadratic Equation