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What is the midpoint of the [tex]$x$[/tex]-intercepts of [tex]$f(x)=(x-2)(x-4)$[/tex]?

A. [tex]$(-3,0)$[/tex]
B. [tex]$(-1,0)$[/tex]
C. [tex]$(1,0)$[/tex]
D. [tex]$(3,0)$[/tex]


Sagot :

To find the midpoint of the [tex]\( x \)[/tex]-intercepts of the function [tex]\( f(x) = (x-2)(x-4) \)[/tex], follow these steps:

1. Identify the [tex]\( x \)[/tex]-intercepts of the function:
The [tex]\( x \)[/tex]-intercepts occur where the function [tex]\( f(x) \)[/tex] equals zero. Set [tex]\( f(x) = 0 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[ (x-2)(x-4) = 0 \][/tex]
This equation is satisfied when either [tex]\( x-2 = 0 \)[/tex] or [tex]\( x-4 = 0 \)[/tex].

Solving these equations:
[tex]\[ x - 2 = 0 \implies x = 2 \][/tex]
[tex]\[ x - 4 = 0 \implies x = 4 \][/tex]

So, the [tex]\( x \)[/tex]-intercepts are [tex]\( x = 2 \)[/tex] and [tex]\( x = 4 \)[/tex].

2. Find the midpoint of these intercepts:
The midpoint of two points [tex]\( (x_1, 0) \)[/tex] and [tex]\( (x_2, 0) \)[/tex] can be found using the midpoint formula for the [tex]\( x \)[/tex]-coordinates:
[tex]\[ \text{Midpoint} = \left( \frac{x_1 + x_2}{2}, 0 \right) \][/tex]

Substitute the values of [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex]:
[tex]\[ x_1 = 2, \quad x_2 = 4 \][/tex]
[tex]\[ \text{Midpoint} = \left( \frac{2 + 4}{2}, 0 \right) = \left( \frac{6}{2}, 0 \right) = (3, 0) \][/tex]

Therefore, the midpoint of the [tex]\( x \)[/tex]-intercepts of the function [tex]\( f(x) = (x-2)(x-4) \)[/tex] is [tex]\( \boxed{(3, 0)} \)[/tex].