Answer :
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.
