1 x² + 3x 3 identify the axis of symmetry and the vertex of the corresponding parabola. Show Bring the quadratic function f(x) 5 into the standard form and your work!
1 x² + 3x 3 identify the axis of symmetry and the vertex of the corresponding parabola. Show Bring the quadratic function f(x) 5 into the standard form and your work!
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![### Quadratic Functions and Parabolas
#### Problem 2
**(a)** Bring the quadratic function \( f(x) = -\frac{1}{3}x^2 + 3x - 5 \) into the standard form and identify the axis of symmetry and the vertex of the corresponding parabola. Show your work!
**Solution:**
1. Begin with the given quadratic function:
\[ f(x) = -\frac{1}{3}x^2 + 3x - 5 \].
2. To convert the quadratic function into its standard form \( f(x) = a(x-h)^2 + k \), we complete the square.
3. Factor out the coefficient of \( x^2 \) from the first two terms:
\[ f(x) = -\frac{1}{3}(x^2 - 9x) - 5 \].
4. Complete the square inside the parentheses. Take half of the coefficient of \( x \), square it, and add and subtract this value inside the parentheses:
\[ x^2 - 9x \]
\[ = x^2 - 9x + \left(\frac{9}{2}\right)^2 - \left(\frac{9}{2}\right)^2 \]
\[ = x^2 - 9x + \frac{81}{4} - \frac{81}{4} \]
\[ = \left(x - \frac{9}{2}\right)^2 - \frac{81}{4} \]
5. Substitute back into the function and simplify:
\[ f(x) = -\frac{1}{3} \left[\left(x - \frac{9}{2}\right)^2 - \frac{81}{4}\right] - 5 \]
\[ = -\frac{1}{3} \left(x - \frac{9}{2}\right)^2 + \frac{81}{12} - 5 \]
\[ = -\frac{1}{3} \left(x - \frac{9}{2}\right)^2 + \frac{81}{12} - \frac{60}{12} \]
\[ = -\frac{1}{3} \left(x - \frac{9}{2}\right)^2 + \frac](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff2ea4237-947f-4fb7-b8d4-fb858663c4d1%2Fe35bac77-7408-48bc-bddc-ca41020edf35%2F6rb537p.jpeg&w=3840&q=75)
Transcribed Image Text:### Quadratic Functions and Parabolas
#### Problem 2
**(a)** Bring the quadratic function \( f(x) = -\frac{1}{3}x^2 + 3x - 5 \) into the standard form and identify the axis of symmetry and the vertex of the corresponding parabola. Show your work!
**Solution:**
1. Begin with the given quadratic function:
\[ f(x) = -\frac{1}{3}x^2 + 3x - 5 \].
2. To convert the quadratic function into its standard form \( f(x) = a(x-h)^2 + k \), we complete the square.
3. Factor out the coefficient of \( x^2 \) from the first two terms:
\[ f(x) = -\frac{1}{3}(x^2 - 9x) - 5 \].
4. Complete the square inside the parentheses. Take half of the coefficient of \( x \), square it, and add and subtract this value inside the parentheses:
\[ x^2 - 9x \]
\[ = x^2 - 9x + \left(\frac{9}{2}\right)^2 - \left(\frac{9}{2}\right)^2 \]
\[ = x^2 - 9x + \frac{81}{4} - \frac{81}{4} \]
\[ = \left(x - \frac{9}{2}\right)^2 - \frac{81}{4} \]
5. Substitute back into the function and simplify:
\[ f(x) = -\frac{1}{3} \left[\left(x - \frac{9}{2}\right)^2 - \frac{81}{4}\right] - 5 \]
\[ = -\frac{1}{3} \left(x - \frac{9}{2}\right)^2 + \frac{81}{12} - 5 \]
\[ = -\frac{1}{3} \left(x - \frac{9}{2}\right)^2 + \frac{81}{12} - \frac{60}{12} \]
\[ = -\frac{1}{3} \left(x - \frac{9}{2}\right)^2 + \frac
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