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What is the minimum value of the function f(x) = 2x^2 - 8x + 10?
What is the minimum value of the function f(x) = 2x^2 - 8x + 10?
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Q1
What is the minimum value of the function f(x) = 2x^2 - 8x + 10?
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The vertex form gives the minimum at x = 2. f(2) = 2(2^2) - 8(2) + 10 = 6.
Questions & Step-by-step Solutions
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Q
Q: What is the minimum value of the function f(x) = 2x^2 - 8x + 10?
Solution:
The vertex form gives the minimum at x = 2. f(2) = 2(2^2) - 8(2) + 10 = 6.
Steps: 8
Show Steps
Step 1: Identify the function we are working with, 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: Since the coefficient of x^2 (which is 2) is positive, the parabola opens upwards, meaning it has a minimum value.
Step 4: To find the x-coordinate of the vertex (which gives the minimum value), use the formula x = -b/(2a).
Step 5: Substitute the values of a and b into the formula: x = -(-8)/(2*2) = 8/4 = 2.
Step 6: Now, substitute x = 2 back into the function to find the minimum value: f(2) = 2(2^2) - 8(2) + 10.
Step 7: Calculate f(2): f(2) = 2(4) - 16 + 10 = 8 - 16 + 10 = 2.
Step 8: Therefore, the minimum value of the function is 2.
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