find a such that f(x)= ax2-4x+3 has a maximum value of 12
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To find the value of "a" such that the function f(x) = ax^2 - 4x + 3 has a maximum value of 12, we need to use the concept of the vertex of a quadratic function.
The vertex of a quadratic function in the form f(x) = ax^2 + bx + c is given by (-b/2a, f(-b/2a)). In this case, we want the maximum value of the function, so the y-coordinate of the vertex should be 12. Therefore, we have:
f(-b/2a) = 12
The quadratic function you provided is f(x) = ax^2 - 4x + 3. To use the formula for the vertex, we need to determine the values of "a," "b," and "c" in this equation. In this case, a = a, b = -4, and c = 3.
Substituting these values, we have:
f(-(-4)/(2a)) = 12
f(4/(2a)) = 12
f(2/a) = 12
Now, we can substitute the general form of f(x) with the given quadratic function:
a * (2/a)^2 - 4 * (2/a) + 3 = 12
Simplifying the equation further, we have:
4 - 8/a + 3 = 12
7 - 8/a = 12
-8/a = 12 - 7
-8/a = 5
-8 = 5a
a = -8/5
Therefore, the value of "a" that makes the function f(x) = ax^2 - 4x + 3 have a maximum value of 12 is -8/5.