To find the [tex]\(x\)[/tex]- and [tex]\(y\)[/tex]-intercepts of the quadratic function [tex]\( f(x) = -x^2 + 9x - 6 \)[/tex], we will follow these steps:
### Finding the [tex]\(y\)[/tex]-Intercept
The [tex]\(y\)[/tex]-intercept of a function occurs where [tex]\(x = 0\)[/tex]. We substitute [tex]\(x = 0\)[/tex] into the function:
[tex]\[ f(0) = -0^2 + 9(0) - 6. \][/tex]
Simplifying this, we get:
[tex]\[ f(0) = -6. \][/tex]
Thus, the [tex]\(y\)[/tex]-intercept is at the point [tex]\((0, -6)\)[/tex].
### Finding the [tex]\(x\)[/tex]-Intercepts
The [tex]\(x\)[/tex]-intercepts are the points where the function [tex]\( f(x) = -x^2 + 9x - 6 \)[/tex] equals zero. We solve the equation:
[tex]\[ -x^2 + 9x - 6 = 0. \][/tex]
This is a quadratic equation, which we can solve using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = -1 \)[/tex], [tex]\( b = 9 \)[/tex], and [tex]\( c = -6 \)[/tex]:
1. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 9^2 - 4(-1)(-6) = 81 - 24 = 57. \][/tex]
2. Compute the roots using the quadratic formula:
[tex]\[ x = \frac{-9 \pm \sqrt{57}}{2(-1)}. \][/tex]
[tex]\[ x = \frac{-9 \pm \sqrt{57}}{-2}. \][/tex]
This results in:
[tex]\[ x = \frac{9 \pm \sqrt{57}}{2}. \][/tex]
Calculating these values, we get two [tex]\(x\)[/tex]-intercepts:
- For the [tex]\(+\sqrt{57}\)[/tex]:
[tex]\[ x = \frac{9 + \sqrt{57}}{2} \approx 8.275. \][/tex]
- For the [tex]\(-\sqrt{57}\)[/tex]:
[tex]\[ x = \frac{9 - \sqrt{57}}{2} \approx 0.725. \][/tex]
Thus, the [tex]\(x\)[/tex]-intercepts, rounded to three decimal places, are approximately at (8.275, 0) and (0.725, 0).
### Summary
- The [tex]\( y \)[/tex]-intercept is at [tex]\((0, -6)\)[/tex].
- The [tex]\( x \)[/tex]-intercepts are at:
[tex]\((8.275, 0)\)[/tex] and [tex]\((0.725, 0)\)[/tex].
So, the completed answer is:
The [tex]\(y\)[/tex]-intercept is at: [tex]\((0, -6)\)[/tex]
The [tex]\(x\)[/tex]-intercepts are at:
[tex]\[(8.275, 0)\][/tex]
[tex]\[(0.725, 0)\][/tex]
