Question

How do you find the domain, VA, HA, Zeros and intercepts of $y = \dfrac{{{x^2} - 16}}{{{x^2} - 4}}?$

Answer 1

According to the question we have to find the domain, VA, HA, Zeros and intercepts of $y = \dfrac{{{x^2} - 16}}{{{x^2} - 4}}$. First of all we have to equating the denominator to zero because the denominator of given expression $y = \dfrac{{{x^2} - 16}}{{{x^2} - 4}}$ cannot be zero as this would make y undefined.
Now, we have to solve the values when the denominator equals to zero and if the numerator is non-zero for these values then they are vertical asymptotes.
Now, we have to find the domain of the given expression in the question and then find the horizontal asymptotes occurs as $\mathop {\lim }\limits_{x \to \pm \infty } ,y \to c$(a constant)
Now, we have to find the intercepts of the given expression $y = \dfrac{{{x^2} - 16}}{{{x^2} - 4}}$as x-intercept and y-intercept, if we calculate y-intercept then x tends to zero and when we calculate x-intercept then y tends to zero.
Finally, we have to find the zeros of the given expression$y = \dfrac{{{x^2} - 16}}{{{x^2} - 4}}$.

Complete step-by-step answer:
Step 1: first of all we have to find the vertical asymptotes as equating the denominator of the given expression $y = \dfrac{{{x^2} - 16}}{{{x^2} - 4}}$ to 0.
$\Rightarrow \left( {{x^2} - 4} \right) = 0$
Now, we know that $\left( {{a^2} - {b^2}} \right)$is equals to $\left( {a - b} \right)\left( {a + b} \right)$
$\Rightarrow \left( {{x^2} - {2^2}} \right) = 0 \\\ \Rightarrow \left( {x - 2} \right)\left( {x + 2} \right) = 0 \\\ \Rightarrow x = + 2, - 2 \\\$
Step 2: now, we have to calculate the domain as we can see that at the values of $x$ which is $+ 2, - 2$the value of given expression $y = \dfrac{{{x^2} - 16}}{{{x^2} - 4}}$is undefined.
So, we have the domain as mentioned below.
Domain $= x \in R,x \ne \pm 2$
Step 3: Now, divide terms on numerator/denominator by the highest power of $x$ that is ${x^2}$

$$\Rightarrow y = \dfrac{{\dfrac{{{x^2}}}{{{x^2}}} - \dfrac{{16}}{{{x^2}}}}}{{\dfrac{{{x^2}}}{{{x^2}}} - \dfrac{4}{{{x^2}}}}} \\\ \Rightarrow y = \dfrac{{1 - \dfrac{{16}}{{{x^2}}}}}{{1 - \dfrac{4}{{{x^2}}}}} \\\$$

Now, we have to find the horizontal asymptote that occurs as $\mathop {\lim }\limits_{x \to \pm \infty } ,y \to c$(a constant).

$$\Rightarrow \mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{1 - \dfrac{{16}}{{{x^2}}}}}{{1 - \dfrac{4}{{{x^2}}}}} \\\ \Rightarrow \dfrac{{1 - \dfrac{{16}}{{{{\left( { \pm \infty } \right)}^2}}}}}{{1 - \dfrac{4}{{{{\left( { \pm \infty } \right)}^2}}}}} \\\$$

We know that $\dfrac{1}{{ \pm \infty }}$tends to 0,

$$\Rightarrow \dfrac{{1 - 0}}{{1 - 0}} \\\ \Rightarrow y = 1 \\\$$

Step 4: Now, we have to find the intercepts of the given expression $y = \dfrac{{{x^2} - 16}}{{{x^2} - 4}}$as x-intercept and y-intercept, if we calculate y-intercept then x tends to zero.
Y-intercept $= x \to 0\left( {\dfrac{{{x^2} - 16}}{{{x^2} - 4}}} \right)$
Y-intercept $= \dfrac{{0 - 16}}{{0 - 4}}$
$\Rightarrow \dfrac{{ - 16}}{{ - 4}} \\\ \Rightarrow 4 \\\$
Step 5: Now, we have to calculate x-intercept then y tends to zero.
X-intercept $= y \to 0\left( {y = \dfrac{{{x^2} - 16}}{{{x^2} - 4}}} \right)$
$\Rightarrow 0 = \dfrac{{{x^2} - 16}}{{{x^2} - 4}} \\\ \Rightarrow 0 = {x^2} - 16 \\\ \Rightarrow 0 = {x^2} - {4^2} \\\ \Rightarrow 0 = \left( {x - 4} \right)\left( {x + 4} \right) \\\ \Rightarrow x = + 4, - 4 \\\$
Step 6: Now, we have to find the zeros of the given expression$y = \dfrac{{{x^2} - 16}}{{{x^2} - 4}}$.
\Rightarrow \left[ {\left( {\dfrac{{{x^2} - 16}}{{{x^2} - 4}}} \right)\left\\{ { - 10,10, - 5,5} \right\\}} \right]

Final solution: The domain of the given expression in the question is $x \in R,x \ne \pm 2$, the VA of the given expression is $x = \pm 2$, the HA of the given expression is y-intercept is 1 and x-intercept is $\pm 4$ and the zeros of the given expression is $\left( { - 10,10, - 5,5} \right)$.

Note:
It is necessary to equivalent the denominator to 0 because at denominator tends 0 the value of the given expression $y = \dfrac{{{x^2} - 16}}{{{x^2} - 4}}$ is undefined.
It is necessary to calculate y-intercept then x tends to zero and when we calculate x-intercept then y tends to zero.