Question

question
find all holes of the following function. if a hole exists, write your answer as a coordinate point in simplest form.
f(x)=\frac{3x^{2}-147}{5x^{2}-36x + 7}
answer
attempt 1 out of 2
add a hole
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Explanation:

Step1: Factor the numerator and denominator

First, factor the numerator $5x^{2}-36x + 7=(5x - 1)(x - 7)$.
Factor the denominator $3x^{2}-147 = 3(x^{2}-49)=3(x + 7)(x - 7)$.

Step2: Identify common factors

The common factor between the numerator and denominator is $(x - 7)$.

Step3: Find the x - value of the hole

Set the common factor equal to zero: $x-7 = 0$, so $x = 7$.

Step4: Find the y - value of the hole

Substitute $x = 7$ into the simplified function. After canceling out the $(x - 7)$ terms, the function becomes $f(x)=\frac{5x - 1}{3(x + 7)}$.
Substitute $x = 7$ into $f(x)=\frac{5x - 1}{3(x + 7)}$, we get $f(7)=\frac{5\times7-1}{3\times(7 + 7)}=\frac{35 - 1}{3\times14}=\frac{34}{42}=\frac{17}{21}$.

Answer:

$(7,\frac{17}{21})$