### Part A: Finding the Points of Intersection
To determine where the paths of the two basketballs intersect, we need to find the solutions to the equation [tex]\( g(x) = h(x) \)[/tex].
Given:
[tex]\[ g(x) = -2x^2 + 13x + 7 \][/tex]
[tex]\[ h(x) = -x^2 + 4x + 21 \][/tex]
Equate the two equations:
[tex]\[ -2x^2 + 13x + 7 = -x^2 + 4x + 21 \][/tex]
Rearrange to set the equation to zero:
[tex]\[ -2x^2 + 13x + 7 + x^2 - 4x - 21 = 0 \][/tex]
[tex]\[ -x^2 + 9x - 14 = 0 \][/tex]
To solve this quadratic equation, we can factor it:
[tex]\[ -x^2 + 9x - 14 = 0 \][/tex]
[tex]\[ x^2 - 9x + 14 = 0 \][/tex]
Factor the quadratic expression:
[tex]\[ (x - 2)(x - 7) = 0 \][/tex]
Set each factor to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x - 2 = 0 \implies x = 2 \][/tex]
[tex]\[ x - 7 = 0 \implies x = 7 \][/tex]
Thus, the distances in meters where the two basketball paths intersect are at [tex]\( x = 2 \)[/tex] meters and [tex]\( x = 7 \)[/tex] meters.
### Part B: Simplifying [tex]\( f(x) = \frac{g(x)}{n(x)} \)[/tex] and Finding Asymptotes and Holes
Given [tex]\( f(x) = \frac{g(x)}{h(x)} \)[/tex]:
[tex]\[ g(x) = -2x^2 + 13x + 7 \][/tex]
[tex]\[ h(x) = -x^2 + 4x + 21 \][/tex]
First, simplify [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = \frac{-2x^2 + 13x + 7}{-x^2 + 4x + 21} \][/tex]
To simplify [tex]\( f(x) \)[/tex], we perform polynomial long division:
After simplification:
[tex]\[ f(x) = \frac{2x + 1}{x + 3} \][/tex]
To identify vertical asymptotes, we find the values of [tex]\( x \)[/tex] for which the denominator is zero:
[tex]\[ x + 3 = 0 \implies x = -3 \][/tex]
[tex]\[ x - 7 = 0 \implies x = 7 \][/tex]
So, the vertical asymptotes are at [tex]\( x = -3 \)[/tex] and [tex]\( x = 7 \)[/tex].
Next, we identify the horizontal asymptotes by comparing the degrees of the numerator and the denominator. Both [tex]\( g(x) \)[/tex] and [tex]\( h(x) \)[/tex] are quadratic polynomials (degree 2).
The leading coefficients of [tex]\( g(x) \)[/tex] and [tex]\( h(x) \)[/tex] are [tex]\(-2\)[/tex] and [tex]\(-1\)[/tex] respectively. The horizontal asymptote is determined by simplifying the ratio of these leading coefficients:
[tex]\[ y = \frac{-2}{-1} = 2 \][/tex]
So, the horizontal asymptote is at [tex]\( y = 2 \)[/tex].
Lastly, to find holes, we look for common factors in the numerator and the denominator that would cancel out. After factorization, we see that [tex]\( x = 7 \)[/tex] is a common factor in both the numerator and the denominator.
[tex]\[ (x - 7) / (x - 7) \][/tex]
Therefore, we have a hole at [tex]\( x = 7 \)[/tex].
### Summary
#### Part A:
The basketball paths cross at [tex]\( x = 2 \)[/tex] meters and [tex]\( x = 7 \)[/tex] meters.
#### Part B:
The simplified form of [tex]\( f(x) \)[/tex] is:
[tex]\[ f(x) = \frac{2x + 1}{x + 3} \][/tex]
- Vertical asymptotes: [tex]\( x = -3 \)[/tex] and [tex]\( x = 7 \)[/tex]
- Horizontal asymptote: [tex]\( y = 2 \)[/tex]
- Hole: At [tex]\( x = 7 \)[/tex]
