The minimum value of the function f(x) = x^2 - 4x + 6 occurs at x = ? (2020)
Practice Questions
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The minimum value of the function f(x) = x^2 - 4x + 6 occurs at x = ? (2020)
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The vertex of the parabola occurs at x = -b/(2a) = 4/2 = 2. The minimum value is f(2) = 2^2 - 4*2 + 6 = 2.
Questions & Step-by-step Solutions
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Q
Q: The minimum value of the function f(x) = x^2 - 4x + 6 occurs at x = ? (2020)
Solution: The vertex of the parabola occurs at x = -b/(2a) = 4/2 = 2. The minimum value is f(2) = 2^2 - 4*2 + 6 = 2.
Steps: 7
Step 1: Identify the function we are working with, which is f(x) = x^2 - 4x + 6.
Step 2: Recognize that this is a quadratic function in the form of ax^2 + bx + c, where a = 1, b = -4, and c = 6.
Step 3: To find the minimum value of the function, we need to find the vertex of the parabola. 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 = -(-4)/(2*1) = 4/2 = 2.
Step 5: Now that we have the x-coordinate of the vertex (x = 2), we need to find the minimum value of the function by substituting x = 2 back into the function f(x).
