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what is the zero of the function

What Is The Zero Of The Function class=

Sagot :

Answer:

[tex]x=6[/tex] (or 6)

Step-by-step explanation:

Set f(x)=0:

[tex]f(x)=\frac{x^2+x-42}{x^2+3x-28}[/tex]

[tex]0=\frac{x^2+x-42}{x^2+3x-28}[/tex]

[tex]0=x^2+x-42[/tex]

[tex]0=(x+7)(x-6)[/tex]

[tex]x=-7[/tex] or [tex]x=6[/tex]

Now check both solutions:

[tex]f(-7)=\frac{(-7)^2+(-7)-42}{(-7)^2+3(-7)-28}[/tex]

[tex]f(-7)=\frac{49-7-42}{49-21-28}[/tex]

[tex]f(-7)=\frac{42-42}{28-28}[/tex]

[tex]f(-7)=\frac{0}{0}[/tex]

Therefore, [tex]x\neq-7[/tex] because there is a hole at [tex](-7,0)[/tex]. The denominator, in addition, can never be 0, so the function is undefined for [tex]x=-7[/tex].

[tex]f(6)=\frac{(6)^2+(6)-42}{(6)^2+3(6)-28}[/tex]

[tex]f(6)=\frac{36+6-42}{36+18-28}[/tex]

[tex]f(6)=\frac{42-42}{54-28}[/tex]

[tex]f(6)=\frac{0}{24}[/tex]

[tex]f(6)=0[/tex]

Since [tex]f(6)=0[/tex], then [tex]x=6[/tex] is the only zero of the function.

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