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Rewrite the quadratic function from standard form to vertex form. (6 points)

[tex]\[ f(x) = x^2 + 10x + 37 \][/tex]


Sagot :

To rewrite [tex]\(f(x) = \pi^2 x^2 + 10x + 37\)[/tex] from standard form [tex]\(ax^2 + bx + c\)[/tex] to vertex form [tex]\(a(x-h)^2 + k\)[/tex], we will use the method known as "completing the square". Here’s the detailed step-by-step process:

1. Identify coefficients:
In the quadratic function [tex]\(f(x) = \pi^2 x^2 + 10x + 37\)[/tex], identify the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex].
- [tex]\(a = \pi^2\)[/tex]
- [tex]\(b = 10\)[/tex]
- [tex]\(c = 37\)[/tex]

2. Factor out [tex]\(a\)[/tex] from the [tex]\(x^2\)[/tex] and [tex]\(x\)[/tex] terms:
Factor out [tex]\(\pi^2\)[/tex] from the first two terms inside the function:
[tex]\[ f(x) = \pi^2 (x^2 + \frac{10}{\pi^2} x) + 37 \][/tex]

3. Complete the square:
To complete the square inside the parentheses, take half of the coefficient of [tex]\(x\)[/tex], square it, and add and subtract it inside the parentheses.
- The coefficient of [tex]\(x\)[/tex] inside the parentheses [tex]\(= \frac{10}{\pi^2}\)[/tex]
- Half of this coefficient [tex]\(= \frac{10}{2\pi^2} = \frac{5}{\pi^2}\)[/tex]
- Square it: [tex]\(\left(\frac{5}{\pi^2}\right)^2 = \frac{25}{\pi^4}\)[/tex]

Now add and subtract this square term inside the parentheses:
[tex]\[ f(x) = \pi^2 \left(x^2 + \frac{10}{\pi^2} x + \frac{25}{\pi^4} - \frac{25}{\pi^4}\right) + 37 \][/tex]

4. Rewrite the perfect square trinomial:
Recognize that [tex]\(x^2 + \frac{10}{\pi^2} x + \frac{25}{\pi^4}\)[/tex] is a perfect square trinomial:
[tex]\[ f(x) = \pi^2 \left(\left( x + \frac{5}{\pi^2} \right)^2 - \frac{25}{\pi^4} \right) + 37 \][/tex]

5. Simplify inside the parentheses:
Distribute [tex]\(\pi^2\)[/tex] and combine the constant terms:
[tex]\[ f(x) = \pi^2 \left( x + \frac{5}{\pi^2} \right)^2 - \pi^2 \cdot \frac{25}{\pi^4} + 37 \][/tex]
Simplify the constant term:
[tex]\[ \pi^2 \cdot \frac{25}{\pi^4} = \frac{25}{\pi^2} \][/tex]

6. Combine the constants:
[tex]\[ f(x) = \pi^2 \left( x + \frac{5}{\pi^2} \right)^2 - \frac{25}{\pi^2} + 37 \][/tex]
Combine the constants:
[tex]\[ f(x) = \pi^2 \left( x + \frac{5}{\pi^2} \right)^2 + \left( 37 - \frac{25}{\pi^2} \right) \][/tex]

So, the quadratic function in vertex form is:
[tex]\[ f(x) = \pi^2 \left( x + \frac{5}{\pi^2} \right)^2 + 37 - \frac{25}{\pi^2} \][/tex]

Here, [tex]\((h, k)\)[/tex] represents the vertex of the parabola, where:
[tex]\( h = -\frac{5}{\pi^2} \)[/tex] and [tex]\( k = 37 - \frac{25}{\pi^2} \)[/tex].