SOLUTION: Let f(x)=x^2+ax+b for some value of a and b. f(x_0) = 0, f(x_1) = 0 and x_0x_1 = 3, x_0 +x_1 = −2. Find the minimum value of the function.
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-> SOLUTION: Let f(x)=x^2+ax+b for some value of a and b. f(x_0) = 0, f(x_1) = 0 and x_0x_1 = 3, x_0 +x_1 = −2. Find the minimum value of the function.
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Question 1152532: Let f(x)=x^2+ax+b for some value of a and b. f(x_0) = 0, f(x_1) = 0 and x_0x_1 = 3, x_0 +x_1 = −2. Find the minimum value of the function.
The reason for these substitutions is to avoid confusion with the 'a' and 'b' in which is a common general form of quadratics used in many math textbooks.
If we compare and , we see that
a = 1
b = p
c = q
Use the quadratic case of Vieta's formulas
Rule: If and are roots to , then and
In this case, and play the role of and (since plugging either x value into f(x) yields f(x) = 0)
Plug in the given roots
Plug in b = p and a = 1.
The sum of the roots is equal to -2 (given).
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Move to the other part of Vieta's formula
Plug in the given roots
Plug in c = q and a = 1.
The product of the roots is 3 (given)
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With p = 2 and q = 3, we can then say,
The equation is in the form with
a = 1
b = 2
c = 3
Find the x coordinate of the vertex
The x coordinate of the vertex is -1. Plug this into the f(x) function to find the y coordinate of the vertex.
The vertex is (h,k) = (-1,2)
Graph of
We see the lowest point is (-1,2) which was the vertex we found earlier.
The final answer is 2 as this is the lowest or smallest output possible of the f(x) function. That's what it means when the teacher wants the minimum value of the function.
Recall that y = f(x) is the output, while x is the input.
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