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What is the minimum value of f(x) = 2x^2 - 8x + 10? (2021)
What is the minimum value of f(x) = 2x^2 - 8x + 10? (2021)
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What is the minimum value of f(x) = 2x^2 - 8x + 10? (2021)
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The minimum occurs at x = -b/(2a) = 8/(2*2) = 2. f(2) = 2(2^2) - 8(2) + 10 = 2.
Questions & Step-by-step Solutions
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Q: What is the minimum value of f(x) = 2x^2 - 8x + 10? (2021)
Solution:
The minimum occurs at x = -b/(2a) = 8/(2*2) = 2. f(2) = 2(2^2) - 8(2) + 10 = 2.
Steps: 9
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Step 1: Identify the function we need to analyze, which is f(x) = 2x^2 - 8x + 10.
Step 2: Recognize that this is a quadratic function in the form of ax^2 + bx + c, where a = 2, b = -8, and c = 10.
Step 3: To find the minimum value of a quadratic function, use the formula x = -b/(2a).
Step 4: Substitute the values of a and b into the formula: x = -(-8)/(2*2).
Step 5: Simplify the expression: x = 8/(4) = 2.
Step 6: Now, we need to find the value of f(x) at x = 2. Substitute x = 2 into the function: f(2) = 2(2^2) - 8(2) + 10.
Step 7: Calculate f(2): f(2) = 2(4) - 16 + 10.
Step 8: Simplify further: f(2) = 8 - 16 + 10 = 2.
Step 9: Therefore, the minimum value of f(x) is 2.
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