Question

How many x-intercepts does the function f(x)=3(2x−1)(x+2)(8x−5)2
have when graphed? (1 point)

Answer 1

To find the number of x-intercepts of the function \( f(x) = 3(2x - 1)(x + 2)(8x - 5)^2 \), we need to determine the values of \( x \) for which \( f(x) = 0 \).

The function can be set equal to zero as follows:

\[
3(2x - 1)(x + 2)(8x - 5)^2 = 0
\]

Since the factor of 3 is a constant and does not affect the roots, we can focus on the polynomial part:

\[
(2x - 1)(x + 2)(8x - 5)^2 = 0
\]

Now we will identify the roots of each factor:

1. From \( 2x - 1 = 0 \):
\[
2x = 1 \implies x = \frac{1}{2}
\]

2. From \( x + 2 = 0 \):
\[
x = -2
\]

3. From \( (8x - 5)^2 = 0 \):
\[
8x - 5 = 0 \implies 8x = 5 \implies x = \frac{5}{8}
\]
The factor \( (8x - 5)^2 \) gives a double root at \( x = \frac{5}{8} \).

Now we summarize the roots:

- \( x = \frac{1}{2} \) (single root)
- \( x = -2 \) (single root)
- \( x = \frac{5}{8} \) (double root)

X-intercepts correspond to the unique values of \( x \) where the function is equal to zero. The roots we have found are:

- \( x = \frac{1}{2} \)
- \( x = -2 \)
- \( x = \frac{5}{8} \)

Counting the unique roots, we have three distinct x-intercepts:

1. \( \frac{1}{2} \)
2. \( -2 \)
3. \( \frac{5}{8} \)

Thus, the function \( f(x) \) has **3 x-intercepts**.