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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![### Finding the x-Intercepts of a Quadratic Function
Given the quadratic function:
\[ f(x) = x^2 + 4x + 1 \]
We aim to find the x-intercepts of the graph of \( y = f(x) \).
#### Step-by-Step Solution:
To find the x-intercepts, we set \( y = 0 \) and solve for \( x \):
\[ f(x) = 0 \]
\[ x^2 + 4x + 1 = 0 \]
This quadratic equation can be solved using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For the equation \( x^2 + 4x + 1 = 0 \), the coefficients are:
- \( a = 1 \)
- \( b = 4 \)
- \( c = 1 \)
Substitute these values into the quadratic formula:
\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ x = \frac{-4 \pm \sqrt{16 - 4}}{2} \]
\[ x = \frac{-4 \pm \sqrt{12}}{2} \]
\[ x = \frac{-4 \pm 2\sqrt{3}}{2} \]
\[ x = -2 \pm \sqrt{3} \]
Therefore, the x-intercepts are at:
\[ x = -2 + \sqrt{3} \]
\[ x = -2 - \sqrt{3} \]
On the provided interface, the options presented for selecting the x-intercepts are:
- A textbox for entering the x-intercepts.
- An option for selecting "No solution" if no real x-intercepts exist.
In this case, there are real solutions, and they should be entered in the textbox if available or selected appropriately.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4f8a397f-4115-4363-8e32-37a2c2950f08%2Fe3f9bf00-89bf-403c-ac76-c1ab06220da1%2Fipz8h1r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Finding the x-Intercepts of a Quadratic Function
Given the quadratic function:
\[ f(x) = x^2 + 4x + 1 \]
We aim to find the x-intercepts of the graph of \( y = f(x) \).
#### Step-by-Step Solution:
To find the x-intercepts, we set \( y = 0 \) and solve for \( x \):
\[ f(x) = 0 \]
\[ x^2 + 4x + 1 = 0 \]
This quadratic equation can be solved using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For the equation \( x^2 + 4x + 1 = 0 \), the coefficients are:
- \( a = 1 \)
- \( b = 4 \)
- \( c = 1 \)
Substitute these values into the quadratic formula:
\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ x = \frac{-4 \pm \sqrt{16 - 4}}{2} \]
\[ x = \frac{-4 \pm \sqrt{12}}{2} \]
\[ x = \frac{-4 \pm 2\sqrt{3}}{2} \]
\[ x = -2 \pm \sqrt{3} \]
Therefore, the x-intercepts are at:
\[ x = -2 + \sqrt{3} \]
\[ x = -2 - \sqrt{3} \]
On the provided interface, the options presented for selecting the x-intercepts are:
- A textbox for entering the x-intercepts.
- An option for selecting "No solution" if no real x-intercepts exist.
In this case, there are real solutions, and they should be entered in the textbox if available or selected appropriately.
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