Question

1. consider these functions:

\\( h(n) = \frac{1}{3}n - 5 \\)
\\( f(x) = h(n)x^2 + 4h(n)x \\)
\\( n \
eq 15 \\)
a) determine the zeros of \\( f(x) \\) when \\( n = 6 \\).
b) determine the values of \\( n \\) for which \\( f(x) \\) has a maximum value. show your work.
2 points (a.rei.a.1)

Explanation:

Part (a)

Step 1: Find \( h(6) \)

First, substitute \( n = 6 \) into \( h(n)=\frac{1}{3}n - 5 \).
\( h(6)=\frac{1}{3}(6)-5 = 2 - 5=-3 \)

Step 2: Substitute \( h(6) \) into \( f(x) \)

Now, substitute \( h(6)=-3 \) into \( f(x)=h(n)x^{2}+4h(n)x \). So \( f(x)=-3x^{2}+4(-3)x=-3x^{2}-12x \)

Step 3: Find the zeros of \( f(x) \)

To find the zeros, set \( f(x) = 0 \):
\( -3x^{2}-12x = 0 \)
Factor out \( -3x \):
\( -3x(x + 4)=0 \)
Set each factor equal to zero:
\( -3x = 0 \) or \( x + 4 = 0 \)
Solving \( -3x = 0 \) gives \( x = 0 \)
Solving \( x + 4 = 0 \) gives \( x=-4 \)

Step 1: Recall the condition for a quadratic to have a maximum

A quadratic function \( f(x)=ax^{2}+bx + c \) has a maximum value if the coefficient of \( x^{2} \) (i.e., \( a \)) is negative. In our function \( f(x)=h(n)x^{2}+4h(n)x \), \( a = h(n) \) and \( b = 4h(n) \)

Step 2: Analyze \( h(n) \)

We know that \( h(n)=\frac{1}{3}n - 5 \)
For \( f(x) \) to have a maximum, we need \( h(n)<0 \)
So, set \( \frac{1}{3}n - 5<0 \)

Step 3: Solve the inequality for \( n \)

Add 5 to both sides:
\( \frac{1}{3}n<5 \)
Multiply both sides by 3:
\( n < 15 \)
We also know from the problem that \( n
eq15 \), so combining these, the values of \( n \) for which \( f(x) \) has a maximum value are all real numbers \( n \) such that \( n < 15 \)

Answer:

The zeros of \( f(x) \) when \( n = 6 \) are \( x = 0 \) and \( x=-4 \)

Part (b)