What is the minimum value of f(x) = x^2 - 6x + 10? (2020)
Practice Questions
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What is the minimum value of f(x) = x^2 - 6x + 10? (2020)
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The vertex form gives the minimum at x = 3. f(3) = 3^2 - 6(3) + 10 = 1.
Questions & Step-by-step Solutions
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Q
Q: What is the minimum value of f(x) = x^2 - 6x + 10? (2020)
Solution: The vertex form gives the minimum at x = 3. f(3) = 3^2 - 6(3) + 10 = 1.
Steps: 7
Step 1: Identify the function we need to analyze, which is f(x) = x^2 - 6x + 10.
Step 2: Recognize that this is a quadratic function in the standard form ax^2 + bx + c, where a = 1, b = -6, and c = 10.
Step 3: To find the minimum value, we can convert the function into vertex form. The x-coordinate of the vertex can be found using the formula x = -b/(2a).
Step 4: Substitute the values of a and b into the formula: x = -(-6)/(2*1) = 6/2 = 3.
Step 5: Now that we have x = 3, we need to find the corresponding y-value (minimum value of f) by substituting x back into the function: f(3) = 3^2 - 6(3) + 10.
Step 6: Calculate f(3): f(3) = 9 - 18 + 10 = 1.
Step 7: Therefore, the minimum value of f(x) is 1.
