Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Get quick and reliable solutions to your questions from a community of experienced professionals on our platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
Alright, let's work through the given function [tex]\( f(x) = \frac{x^2 - 6x + 8}{x + 2} \)[/tex].
1. Domain of the function:
To find the domain of the function, we need to determine where the denominator is zero because division by zero is undefined.
Set the denominator equal to zero:
[tex]\[ x + 2 = 0 \][/tex]
[tex]\[ x = -2 \][/tex]
So, the function [tex]\( f(x) \)[/tex] is undefined at [tex]\( x = -2 \)[/tex]. Therefore, the domain of the function in interval notation is:
[tex]\[ (-\infty, -2) \cup (-2, \infty) \][/tex]
2. X-intercepts:
To find the x-intercepts, we need to set the numerator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 - 6x + 8 = 0 \][/tex]
This is a quadratic equation, and solving it, we get:
[tex]\[ (x-2)(x-4) = 0 \][/tex]
[tex]\[ x = 2 \quad \text{and} \quad x = 4 \][/tex]
Therefore, the x-intercepts are:
[tex]\[ (2, 0) \quad \text{and} \quad (4, 0) \][/tex]
3. Y-intercept:
To find the y-intercept, we evaluate the function [tex]\( f(x) \)[/tex] at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{0^2 - 6 \cdot 0 + 8}{0 + 2} = \frac{8}{2} = 4 \][/tex]
So, the y-intercept is:
[tex]\[ (0, 4) \][/tex]
Final Answers:
- The domain of the function is:
[tex]\[ (-\infty, -2) \cup (-2, \infty) \][/tex]
- The x-intercepts are:
[tex]\[ (2, 0) \quad \text{and} \quad (4, 0) \][/tex]
- The y-intercept is:
[tex]\[ (0, 4) \][/tex]
1. Domain of the function:
To find the domain of the function, we need to determine where the denominator is zero because division by zero is undefined.
Set the denominator equal to zero:
[tex]\[ x + 2 = 0 \][/tex]
[tex]\[ x = -2 \][/tex]
So, the function [tex]\( f(x) \)[/tex] is undefined at [tex]\( x = -2 \)[/tex]. Therefore, the domain of the function in interval notation is:
[tex]\[ (-\infty, -2) \cup (-2, \infty) \][/tex]
2. X-intercepts:
To find the x-intercepts, we need to set the numerator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 - 6x + 8 = 0 \][/tex]
This is a quadratic equation, and solving it, we get:
[tex]\[ (x-2)(x-4) = 0 \][/tex]
[tex]\[ x = 2 \quad \text{and} \quad x = 4 \][/tex]
Therefore, the x-intercepts are:
[tex]\[ (2, 0) \quad \text{and} \quad (4, 0) \][/tex]
3. Y-intercept:
To find the y-intercept, we evaluate the function [tex]\( f(x) \)[/tex] at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{0^2 - 6 \cdot 0 + 8}{0 + 2} = \frac{8}{2} = 4 \][/tex]
So, the y-intercept is:
[tex]\[ (0, 4) \][/tex]
Final Answers:
- The domain of the function is:
[tex]\[ (-\infty, -2) \cup (-2, \infty) \][/tex]
- The x-intercepts are:
[tex]\[ (2, 0) \quad \text{and} \quad (4, 0) \][/tex]
- The y-intercept is:
[tex]\[ (0, 4) \][/tex]
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.