IDNLearn.com offers a unique blend of expert answers and community insights. Get the information you need quickly and accurately with our reliable and thorough Q&A platform.
Sagot :
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].
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].
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Thank you for choosing IDNLearn.com. We’re committed to providing accurate answers, so visit us again soon.