Connect with a knowledgeable community and get your questions answered on IDNLearn.com. Ask your questions and receive detailed and reliable answers from our experienced and knowledgeable community members.
Sagot :
To determine the [tex]$x$[/tex]-intercepts of the function [tex]$f(x) = x^2 - 25$[/tex], we need to find the values of [tex]$x$[/tex] where [tex]$f(x) = 0$[/tex]. This involves solving the equation:
[tex]\[ f(x) = 0 \][/tex]
Given our function:
[tex]\[ x^2 - 25 = 0 \][/tex]
To solve for [tex]$x$[/tex], we first isolate the [tex]$x^2$[/tex] term:
[tex]\[ x^2 = 25 \][/tex]
Next, we take the square root of both sides of the equation to solve for [tex]$x$[/tex]:
[tex]\[ x = \pm \sqrt{25} \][/tex]
This simplifies to:
[tex]\[ x = \pm 5 \][/tex]
Therefore, the [tex]$x$[/tex]-intercepts of the function are [tex]$x = 5$[/tex] and [tex]$x = -5$[/tex].
Given the options:
A. -15
B. -20
C. -5
D. -25
We can see that the only [tex]$x$[/tex]-intercept from the given options is:
C. -5
Thus, the correct answer is C. -5.
[tex]\[ f(x) = 0 \][/tex]
Given our function:
[tex]\[ x^2 - 25 = 0 \][/tex]
To solve for [tex]$x$[/tex], we first isolate the [tex]$x^2$[/tex] term:
[tex]\[ x^2 = 25 \][/tex]
Next, we take the square root of both sides of the equation to solve for [tex]$x$[/tex]:
[tex]\[ x = \pm \sqrt{25} \][/tex]
This simplifies to:
[tex]\[ x = \pm 5 \][/tex]
Therefore, the [tex]$x$[/tex]-intercepts of the function are [tex]$x = 5$[/tex] and [tex]$x = -5$[/tex].
Given the options:
A. -15
B. -20
C. -5
D. -25
We can see that the only [tex]$x$[/tex]-intercept from the given options is:
C. -5
Thus, the correct answer is C. -5.
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. IDNLearn.com is your reliable source for answers. We appreciate your visit and look forward to assisting you again soon.