From tech troubles to travel tips, IDNLearn.com has answers to all your questions. Our experts provide timely and accurate responses to help you navigate any topic or issue with confidence.
Sagot :
Let's analyze the function [tex]\( f(x) = 2x^2 - x - 6 \)[/tex] and determine which statements about its graph are true. We'll do this step-by-step:
1. Find the Vertex:
The vertex form of a quadratic function [tex]\( ax^2 + bx + c \)[/tex] is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = 2 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -6 \)[/tex].
Calculating the x-coordinate of the vertex:
[tex]\[ x = -\frac{-1}{2 \times 2} = \frac{1}{4} \][/tex]
Now, we substitute [tex]\( x = \frac{1}{4} \)[/tex] back into the function to find the y-coordinate:
[tex]\[ f\left(\frac{1}{4}\right) = 2\left(\frac{1}{4}\right)^2 - \frac{1}{4} - 6 = 2\left(\frac{1}{16}\right) - \frac{1}{4} - 6 = \frac{2}{16} - \frac{1}{4} - 6 = \frac{1}{8} - \frac{2}{8} - 6 = -\frac{1}{8} - 6 = -6\frac{1}{8} \][/tex]
Therefore, the vertex of the function is:
[tex]\[ \left( \frac{1}{4}, -6\frac{1}{8} \right) \][/tex]
2. Domain of the Function:
The domain of any quadratic function [tex]\( ax^2 + bx + c \)[/tex] is all real numbers, i.e.,
[tex]\[ \mathbb{R} = (-\infty, \infty) \][/tex]
So, the statement "The domain of the function is [tex]\( \left\{ x \left| x \geq \frac{1}{4} \right. \right\} \)[/tex] is incorrect.
3. Range of the Function:
Since the coefficient of [tex]\(x^2\)[/tex] is positive ([tex]\(a = 2\)[/tex]), the parabola opens upwards. Thus, the function has a minimum value at its vertex.
Therefore, the range of the function is:
[tex]\[ \left[ -6\frac{1}{8}, \infty \right) \][/tex]
So, the statement "The range of the function is all real numbers" is incorrect.
4. x-intercepts:
To find the [tex]\(x\)[/tex]-intercepts, we set [tex]\(f(x) = 0\)[/tex] and solve for [tex]\(x\)[/tex]:
[tex]\[ 2x^2 - x - 6 = 0 \][/tex]
Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-6)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 48}}{4} = \frac{1 \pm \sqrt{49}}{4} = \frac{1 \pm 7}{4} \][/tex]
Therefore, we have two [tex]\(x\)[/tex]-intercepts:
[tex]\[ x = \frac{8}{4} = 2 \quad \text{and} \quad x = \frac{-6}{4} = -\frac{3}{2} \][/tex]
Hence, the statement "The function has two [tex]\(x\)[/tex]-intercepts" is correct.
5. Intervals of Increasing and Decreasing:
To find where the function is increasing or decreasing, we look at the first derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = 4x - 1 \][/tex]
Setting the derivative to zero to find critical points:
[tex]\[ 4x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{4} \][/tex]
Since the derivative changes from negative to positive at [tex]\( x = \frac{1}{4} \)[/tex], the function is decreasing for [tex]\( x < \frac{1}{4} \)[/tex] and increasing for [tex]\( x > \frac{1}{4} \)[/tex].
Next, the function starts increasing from the vertex [tex]\( y = -6\frac{1}{8} \)[/tex]. Thus, the function is increasing over the interval [tex]\( \left( -6\frac{1}{8}, \infty \right) \)[/tex] is incorrect – it should be [tex]\(\left( \frac{1}{4}, \infty \right)\)[/tex] instead.
In summary:
- The vertex statement is correct.
- The x-intercepts statement is correct.
Thus, the correct statements about the graph of the function [tex]\( f(x) = 2x^2 - x - 6 \)[/tex] are:
- The vertex of the function is [tex]\( \left( \frac{1}{4}, -6\frac{1}{8} \right) \)[/tex].
- The function has two [tex]\( x \)[/tex]-intercepts.
1. Find the Vertex:
The vertex form of a quadratic function [tex]\( ax^2 + bx + c \)[/tex] is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = 2 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -6 \)[/tex].
Calculating the x-coordinate of the vertex:
[tex]\[ x = -\frac{-1}{2 \times 2} = \frac{1}{4} \][/tex]
Now, we substitute [tex]\( x = \frac{1}{4} \)[/tex] back into the function to find the y-coordinate:
[tex]\[ f\left(\frac{1}{4}\right) = 2\left(\frac{1}{4}\right)^2 - \frac{1}{4} - 6 = 2\left(\frac{1}{16}\right) - \frac{1}{4} - 6 = \frac{2}{16} - \frac{1}{4} - 6 = \frac{1}{8} - \frac{2}{8} - 6 = -\frac{1}{8} - 6 = -6\frac{1}{8} \][/tex]
Therefore, the vertex of the function is:
[tex]\[ \left( \frac{1}{4}, -6\frac{1}{8} \right) \][/tex]
2. Domain of the Function:
The domain of any quadratic function [tex]\( ax^2 + bx + c \)[/tex] is all real numbers, i.e.,
[tex]\[ \mathbb{R} = (-\infty, \infty) \][/tex]
So, the statement "The domain of the function is [tex]\( \left\{ x \left| x \geq \frac{1}{4} \right. \right\} \)[/tex] is incorrect.
3. Range of the Function:
Since the coefficient of [tex]\(x^2\)[/tex] is positive ([tex]\(a = 2\)[/tex]), the parabola opens upwards. Thus, the function has a minimum value at its vertex.
Therefore, the range of the function is:
[tex]\[ \left[ -6\frac{1}{8}, \infty \right) \][/tex]
So, the statement "The range of the function is all real numbers" is incorrect.
4. x-intercepts:
To find the [tex]\(x\)[/tex]-intercepts, we set [tex]\(f(x) = 0\)[/tex] and solve for [tex]\(x\)[/tex]:
[tex]\[ 2x^2 - x - 6 = 0 \][/tex]
Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-6)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 48}}{4} = \frac{1 \pm \sqrt{49}}{4} = \frac{1 \pm 7}{4} \][/tex]
Therefore, we have two [tex]\(x\)[/tex]-intercepts:
[tex]\[ x = \frac{8}{4} = 2 \quad \text{and} \quad x = \frac{-6}{4} = -\frac{3}{2} \][/tex]
Hence, the statement "The function has two [tex]\(x\)[/tex]-intercepts" is correct.
5. Intervals of Increasing and Decreasing:
To find where the function is increasing or decreasing, we look at the first derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = 4x - 1 \][/tex]
Setting the derivative to zero to find critical points:
[tex]\[ 4x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{4} \][/tex]
Since the derivative changes from negative to positive at [tex]\( x = \frac{1}{4} \)[/tex], the function is decreasing for [tex]\( x < \frac{1}{4} \)[/tex] and increasing for [tex]\( x > \frac{1}{4} \)[/tex].
Next, the function starts increasing from the vertex [tex]\( y = -6\frac{1}{8} \)[/tex]. Thus, the function is increasing over the interval [tex]\( \left( -6\frac{1}{8}, \infty \right) \)[/tex] is incorrect – it should be [tex]\(\left( \frac{1}{4}, \infty \right)\)[/tex] instead.
In summary:
- The vertex statement is correct.
- The x-intercepts statement is correct.
Thus, the correct statements about the graph of the function [tex]\( f(x) = 2x^2 - x - 6 \)[/tex] are:
- The vertex of the function is [tex]\( \left( \frac{1}{4}, -6\frac{1}{8} \right) \)[/tex].
- The function has two [tex]\( x \)[/tex]-intercepts.
Your presence in our community is highly appreciated. Keep sharing your insights and solutions. Together, we can build a rich and valuable knowledge resource for everyone. Thanks for visiting IDNLearn.com. We’re dedicated to providing clear answers, so visit us again for more helpful information.
