Certainly! Let's find the value of [tex]\(a\)[/tex] such that the quadratic function [tex]\(f(x) = a x^2 - 4 x + 3\)[/tex] attains a maximum value of 12.
To determine the value of [tex]\(a\)[/tex], we need to perform the following steps:
1. Identify the vertex of the quadratic function:
For a quadratic function in the form [tex]\(f(x) = a x^2 + b x + c\)[/tex], the x-coordinate of the vertex, [tex]\(x_{\text{vertex}}\)[/tex], is given by:
[tex]\[
x_{\text{vertex}} = -\frac{b}{2a}
\][/tex]
In our function, [tex]\(b = -4\)[/tex] and [tex]\(c = 3\)[/tex].
2. Set the vertex point and substitute into the function:
Since the function [tex]\(f(x)\)[/tex] has a maximum value at its vertex, we know that:
[tex]\[
f\left(-\frac{b}{2a}\right) = 12
\][/tex]
3. Calculate [tex]\(x_{\text{vertex}}\)[/tex]:
We first calculate [tex]\(x_{\text{vertex}}\)[/tex]:
[tex]\[
x_{\text{vertex}} = -\frac{-4}{2a} = \frac{4}{2a} = \frac{2}{a}
\][/tex]
4. Substitute [tex]\(x_{\text{vertex}}\)[/tex] into the function and set it to 12:
Substitute [tex]\(x_{\text{vertex}} = \frac{2}{a}\)[/tex] into the function and set the resulting expression equal to the maximum value, which is 12:
[tex]\[
f\left(\frac{2}{a}\right) = a \left(\frac{2}{a}\right)^2 - 4 \left(\frac{2}{a}\right) + 3 = 12
\][/tex]
Simplify the expression:
[tex]\[
a \left(\frac{4}{a^2}\right) - \frac{8}{a} + 3 = 12
\][/tex]
[tex]\[
\frac{4}{a} - \frac{8}{a} + 3 = 12
\][/tex]
5. Combine the terms:
Combine the terms with [tex]\(a\)[/tex] as the denominator:
[tex]\[
\frac{4 - 8}{a} + 3 = 12
\][/tex]
[tex]\[
\frac{-4}{a} + 3 = 12
\][/tex]
6. Solve for [tex]\(a\)[/tex]:
Isolate the term involving [tex]\(a\)[/tex]:
[tex]\[
\frac{-4}{a} + 3 = 12
\][/tex]
[tex]\[
\frac{-4}{a} = 12 - 3
\][/tex]
[tex]\[
\frac{-4}{a} = 9
\][/tex]
[tex]\[
-4 = 9a
\][/tex]
[tex]\[
a = -\frac{4}{9}
\][/tex]
Therefore, the value of [tex]\(a\)[/tex] that makes the quadratic function [tex]\(f(x) = a x^2 - 4 x + 3\)[/tex] attain a maximum value of 12 is [tex]\(\boxed{-\frac{4}{9}}\)[/tex].
