find a such that f(x)= ax2-4x+3 has a maximum value of 12

To find the value of "a" that gives the quadratic function f(x) = ax^2 - 4x + 3 a maximum value of 12, we can use the vertex formula.

The vertex formula for a quadratic function f(x) = ax^2 + bx + c is given by:
x = -b / (2a)

In this case, since we are looking for a maximum, the coefficient of x^2 (a) must be negative. So, let's assume a = -k, where k is a positive constant.

Substituting it into the vertex formula, we have:
x = -(-4) / (2 * -k)
x = 4 / (2k)
x = 2/k

Now, we need to find the corresponding value of f(x) at this x-coordinate. Substituting x = 2/k into f(x), we get:
f(2/k) = a(2/k)^2 - 4(2/k) + 3
f(2/k) = 4a/k^2 - 8/k + 3

Since we want this maximum value to be 12, we have:
f(2/k) = 12
4a/k^2 - 8/k + 3 = 12
4a/k^2 - 8/k = 12 - 3
4a/k^2 - 8/k = 9

To simplify further, we can multiply each term by k^2 to eliminate the denominator:
4a - 8k = 9k^2

Now, we have a quadratic equation in terms of k. Rearranging the equation, we get:
9k^2 + 8k - 4a = 0

For the function to have a maximum at f(x) = 12, the discriminant (b^2 - 4ac) of the quadratic equation must be zero. So, we have:
(8^2) - 4(9)(-4a) = 0
64 + 144a = 0
144a = -64
a = -64/144
a = -4/9

Therefore, the value of "a" that gives the quadratic function f(x) = ax^2 - 4x + 3 a maximum value of 12 is -4/9.

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 properties of quadratic functions.

The maximum or minimum value of a quadratic function occurs at its vertex. For a quadratic function in the form f(x) = ax^2 + bx + c, the x-coordinate of the vertex is given by:

x = -b/(2a)

In this case, the quadratic function is f(x) = ax^2 - 4x + 3, and we want the maximum value to be 12. So, we set f(x) = 12:

12 = ax^2 - 4x + 3

Now, we rearrange the equation to have it in the standard form, which is a quadratic equation equal to zero:

ax^2 - 4x + 3 - 12 = 0

ax^2 - 4x - 9 = 0

To find the value of 'a', we need to substitute the x-coordinate of the vertex into the equation and solve for 'a'. The x-coordinate of the vertex is given by x = -b/(2a), which, in this case, is x = (-(-4))/(2a) = 4/(2a) = 2/a.

So, we substitute 2/a into the equation:

a(2/a)^2 - 4(2/a) - 9 = 0

4/a - 8/a - 9 = 0

Multiplying through by 'a' to remove the denominators:

4 - 8 - 9a = 0

-9a - 4 = 0

-9a = 4

a = 4/(-9)

a = -4/9

Therefore, the value of 'a' that makes f(x) = ax^2 - 4x + 3 have a maximum value of 12 is a = -4/9.

let's complete the square

f(x) = a[ x^2 -4/a + 4/a^2 - 4/a^2] + 3
= a(x-2/a) -4/a + 3

so -4/a + 3 = 12
-4a = 9
a = -9/4