To find the [tex]\( x \)[/tex]- and [tex]\( y \)[/tex]-intercepts of the quadratic equation [tex]\( y = -x^2 + 8x - 15 \)[/tex], we proceed as follows:
### Finding the [tex]\( x \)[/tex]-Intercepts:
The [tex]\( x \)[/tex]-intercepts occur where the graph intersects the [tex]\( x \)[/tex]-axis. At these points, [tex]\( y = 0 \)[/tex]. So we set the equation equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 0 = -x^2 + 8x - 15. \][/tex]
This is a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = -1 \)[/tex], [tex]\( b = 8 \)[/tex], and [tex]\( c = -15 \)[/tex].
We can solve this equation using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \][/tex]
Plugging in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ x = \frac{-8 \pm \sqrt{8^2 - 4(-1)(-15)}}{2(-1)}. \][/tex]
[tex]\[ x = \frac{-8 \pm \sqrt{64 - 60}}{-2}. \][/tex]
[tex]\[ x = \frac{-8 \pm \sqrt{4}}{-2}. \][/tex]
[tex]\[ x = \frac{-8 \pm 2}{-2}. \][/tex]
This gives us two solutions:
1. [tex]\( x_1 = \frac{-8 + 2}{-2} = \frac{-6}{-2} = 3 \)[/tex].
2. [tex]\( x_2 = \frac{-8 - 2}{-2} = \frac{-10}{-2} = 5 \)[/tex].
So, the [tex]\( x \)[/tex]-intercepts are [tex]\( x = 3 \)[/tex] and [tex]\( x = 5 \)[/tex].
### Finding the [tex]\( y \)[/tex]-Intercept:
The [tex]\( y \)[/tex]-intercept occurs where the graph intersects the [tex]\( y \)[/tex]-axis. At this point, [tex]\( x = 0 \)[/tex]. So we substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = -0^2 + 8(0) - 15. \][/tex]
[tex]\[ y = -15. \][/tex]
So, the [tex]\( y \)[/tex]-intercept is [tex]\( y = -15 \)[/tex].
### Summary:
- The [tex]\( x \)[/tex]-intercepts of the parabola are [tex]\( x = 3 \)[/tex] and [tex]\( x = 5 \)[/tex].
- The [tex]\( y \)[/tex]-intercept of the parabola is [tex]\( y = -15 \)[/tex].
Therefore:
For the [tex]\( x \)[/tex]-intercept:
A. [tex]\( x = 3, 5 \)[/tex]
For the [tex]\( y \)[/tex]-intercept:
y = -15
