Answer :
To determine between which two ordered pairs the graph of the function [tex]\( f(x) = \frac{1}{2} x^2 + x - 9 \)[/tex] crosses the negative [tex]\( x \)[/tex]-axis, we can solve the quadratic equation [tex]\( \frac{1}{2} x^2 + x - 9 = 0 \)[/tex] using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Given the coefficients:
- [tex]\( a = \frac{1}{2} \)[/tex]
- [tex]\( b = 1 \)[/tex]
- [tex]\( c = -9 \)[/tex]
First, we calculate the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac = 1^2 - 4 \left(\frac{1}{2}\right)(-9) = 1 + 18 = 19 \][/tex]
The quadratic formula then gives us the solutions (roots):
[tex]\[ x = \frac{-1 \pm \sqrt{19}}{1} \][/tex]
[tex]\[ x = -1 \pm \sqrt{19} \][/tex]
Thus, the two roots are:
[tex]\[ x_1 = -1 + \sqrt{19} \][/tex]
[tex]\[ x_2 = -1 - \sqrt{19} \][/tex]
Evaluating these roots:
[tex]\[ x_1 \approx -1 + 4.358898943540674 = 3.358898943540674 \][/tex]
[tex]\[ x_2 \approx -1 - 4.358898943540674 = -5.358898943540674 \][/tex]
Since we are interested in where the graph crosses the negative [tex]\( x \)[/tex]-axis, we take the root corresponding to the negative [tex]\( x \)[/tex]-value:
[tex]\[ x_2 \approx -5.358898943540674 \][/tex]
By examining the interval options provided:
- [tex]\((-6, 0)\)[/tex] and [tex]\((-5, 0)\)[/tex]
- [tex]\((-4, 0)\)[/tex] and [tex]\((-3, 0)\)[/tex]
- [tex]\((-3, 0)\)[/tex] and [tex]\((-2, 0)\)[/tex]
- [tex]\((-2, 0)\)[/tex] and [tex]\((-1, 0)\)[/tex]
We see that [tex]\(-5.358898943540674\)[/tex] is approximately between [tex]\((-6, 0)\)[/tex] and [tex]\((-5, 0)\)[/tex]. Thus, the correct answer is:
[tex]\[ (-6, 0) \text{ and } (-5, 0) \][/tex]
So the graph of [tex]\( f(x) = \frac{1}{2} x^2 + x - 9 \)[/tex] crosses the negative [tex]\( x \)[/tex]-axis between [tex]\((-6, 0)\)[/tex] and [tex]\((-5, 0)\)[/tex].
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Given the coefficients:
- [tex]\( a = \frac{1}{2} \)[/tex]
- [tex]\( b = 1 \)[/tex]
- [tex]\( c = -9 \)[/tex]
First, we calculate the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac = 1^2 - 4 \left(\frac{1}{2}\right)(-9) = 1 + 18 = 19 \][/tex]
The quadratic formula then gives us the solutions (roots):
[tex]\[ x = \frac{-1 \pm \sqrt{19}}{1} \][/tex]
[tex]\[ x = -1 \pm \sqrt{19} \][/tex]
Thus, the two roots are:
[tex]\[ x_1 = -1 + \sqrt{19} \][/tex]
[tex]\[ x_2 = -1 - \sqrt{19} \][/tex]
Evaluating these roots:
[tex]\[ x_1 \approx -1 + 4.358898943540674 = 3.358898943540674 \][/tex]
[tex]\[ x_2 \approx -1 - 4.358898943540674 = -5.358898943540674 \][/tex]
Since we are interested in where the graph crosses the negative [tex]\( x \)[/tex]-axis, we take the root corresponding to the negative [tex]\( x \)[/tex]-value:
[tex]\[ x_2 \approx -5.358898943540674 \][/tex]
By examining the interval options provided:
- [tex]\((-6, 0)\)[/tex] and [tex]\((-5, 0)\)[/tex]
- [tex]\((-4, 0)\)[/tex] and [tex]\((-3, 0)\)[/tex]
- [tex]\((-3, 0)\)[/tex] and [tex]\((-2, 0)\)[/tex]
- [tex]\((-2, 0)\)[/tex] and [tex]\((-1, 0)\)[/tex]
We see that [tex]\(-5.358898943540674\)[/tex] is approximately between [tex]\((-6, 0)\)[/tex] and [tex]\((-5, 0)\)[/tex]. Thus, the correct answer is:
[tex]\[ (-6, 0) \text{ and } (-5, 0) \][/tex]
So the graph of [tex]\( f(x) = \frac{1}{2} x^2 + x - 9 \)[/tex] crosses the negative [tex]\( x \)[/tex]-axis between [tex]\((-6, 0)\)[/tex] and [tex]\((-5, 0)\)[/tex].