Math Problem Statement
In parts (i) and (ii) let f(x) = -x^2 + 4x - 2. Complete the following tables of values.
Solution
The problem provided involves filling out tables based on the function .
Let's break it down step by step.
Part (i): Table 1
You need to compute for different values of and fill in the missing values.
Given table:
\hline x & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & & 1 & & & & -2 & \\ \hline \end{array}$$ Let's compute: 1. For $$x = -1$$: $$f(-1) = -(-1)^2 + 4(-1) - 2 = -1 - 4 - 2 = -7$$ 2. For $$x = 1$$: $$f(1) = -(1)^2 + 4(1) - 2 = -1 + 4 - 2 = 1$$ 3. For $$x = 2$$: $$f(2) = -(2)^2 + 4(2) - 2 = -4 + 8 - 2 = 2$$ 4. For $$x = 3$$: $$f(3) = -(3)^2 + 4(3) - 2 = -9 + 12 - 2 = 1$$ 5. For $$x = 5$$: $$f(5) = -(5)^2 + 4(5) - 2 = -25 + 20 - 2 = -7$$ Thus, the completed table becomes: $$\begin{array}{|c|c|c|c|c|c|c|c|} \hline x & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & -7 & 1 & 1 & 2 & 1 & -2 & -7 \\ \hline \end{array}$$ ### Part (ii): Table 2 In this table, we need to compute $$f(x) + 2$$ for different values of $$x$$. Given table: $$\begin{array}{|c|c|c|c|c|c|c|c|} \hline x & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) + 2 & & & & & 3 & & \\ \hline \end{array}$$ Using values from the completed first table: 1. $$f(-1) = -7$$, so $$f(-1) + 2 = -7 + 2 = -5$$. 2. $$f(0) = 1$$, so $$f(0) + 2 = 1 + 2 = 3$$. 3. $$f(1) = 1$$, so $$f(1) + 2 = 1 + 2 = 3$$. 4. $$f(2) = 2$$, so $$f(2) + 2 = 2 + 2 = 4$$. 5. $$f(4) = -2$$, so $$f(4) + 2 = -2 + 2 = 0$$. 6. $$f(5) = -7$$, so $$f(5) + 2 = -7 + 2 = -5$$. Thus, the completed table becomes: $$\begin{array}{|c|c|c|c|c|c|c|c|} \hline x & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) + 2 & -5 & 3 & 3 & 4 & 3 & 0 & -5 \\ \hline \end{array}$$ ### Part (iii): Table 3 In this table, we need to compute $$f(x) - 4$$ for different values of $$x$$. Given table: $$\begin{array}{|c|c|c|c|c|c|c|c|} \hline x & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) - 4 & & & & & & -11 & \\ \hline \end{array}$$ Using values from the completed first table: 1. $$f(-1) = -7$$, so $$f(-1) - 4 = -7 - 4 = -11$$. 2. $$f(0) = 1$$, so $$f(0) - 4 = 1 - 4 = -3$$. 3. $$f(1) = 1$$, so $$f(1) - 4 = 1 - 4 = -3$$. 4. $$f(2) = 2$$, so $$f(2) - 4 = 2 - 4 = -2$$. 5. $$f(3) = 1$$, so $$f(3) - 4 = 1 - 4 = -3$$. 6. $$f(5) = -7$$, so $$f(5) - 4 = -7 - 4 = -11$$. Thus, the completed table becomes: $$\begin{array}{|c|c|c|c|c|c|c|c|} \hline x & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) - 4 & -11 & -3 & -3 & -2 & -3 & -11 & -11 \\ \hline \end{array}$$ --- Would you like further details or have any questions? Here are 5 additional related questions: 1. How do you find the vertex of the quadratic function $$f(x) = -x^2 + 4x - 2$$? 2. What is the range of the function $$f(x) = -x^2 + 4x - 2$$? 3. How do you graph a quadratic function? 4. How does the transformation $$f(x) + k$$ shift the graph of a quadratic function? 5. Can you explain how completing the square can help in understanding the quadratic function? **Tip:** For any quadratic function of the form $$ax^2 + bx + c$$, the axis of symmetry is given by $$x = -\frac{b}{2a}$$, which is helpful in finding the vertex.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Quadratic Functions
Function Evaluation
Algebra
Formulas
f(x) = -x^2 + 4x - 2
Theorems
N/A
Suitable Grade Level
Grades 9-11
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