QuestionFind the holes in the rational function $f(x)=\frac{x^{2}-4}{x+2}$ . If none, state 'none'.

Studdy Solution

STEP 1

Assumptions1. The function is $f(x)=\frac{x^{}-4}{x+}$ . We are asked to find the coordinates of any holes in the function.

STEP 2

A hole in a rational function occurs when a factor in the denominator of the function also appears in the numerator, causing the function to be undefined at that point.To find the holes, we first factorize the numerator and the denominator.
The numerator $x^{2}-4$ is a difference of squares and can be factored as $(x-2)(x+2)$ .
The denominator $x+2$ is already factored.

STEP 3

Now, we write the function in its factored form.
$f(x)=\frac{(x-2)(x+2)}{x+2}$

STEP 4

We can see that the factor $(x+2)$ appears in both the numerator and the denominator. This means that the function is undefined at $x=-2$ . However, because the factor is cancelled out, this point is a hole in the function, not a vertical asymptote.

STEP 5

To find the y-coordinate of the hole, we substitute $x=-2$ into the simplified function $f(x)=x-2$ .
$f(-2)=-2-2$

STEP 6

Calculate the y-coordinate of the hole.
$f(-2)=-2-2=-4$ So, the coordinates of the hole are $(-2, -4)$ .

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QuestionFind the holes in the rational function $f(x)=\frac{x^{2}-4}{x+2}$ . If none, state 'none'.

Studdy Solution

STEP 1

Assumptions1. The function is $f(x)=\frac{x^{}-4}{x+}$ . We are asked to find the coordinates of any holes in the function.

STEP 2

STEP 3

Now, we write the function in its factored form.
$f(x)=\frac{(x-2)(x+2)}{x+2}$

STEP 4

We can see that the factor $(x+2)$ appears in both the numerator and the denominator. This means that the function is undefined at $x=-2$ . However, because the factor is cancelled out, this point is a hole in the function, not a vertical asymptote.

STEP 5

To find the y-coordinate of the hole, we substitute $x=-2$ into the simplified function $f(x)=x-2$ .
$f(-2)=-2-2$

STEP 6

Calculate the y-coordinate of the hole.
$f(-2)=-2-2=-4$ So, the coordinates of the hole are $(-2, -4)$ .

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QuestionFind the holes in the rational function f(x)=x2−4x+2f(x)=\frac{x^{2}-4}{x+2}f(x)=x+2x2−4​. If none, state 'none'.

Studdy Solution

STEP 1

Assumptions1. The function is f(x)=x−4x+f(x)=\frac{x^{}-4}{x+}f(x)=x+x−4​ . We are asked to find the coordinates of any holes in the function.

STEP 2

STEP 3

Now, we write the function in its factored form.f(x)=(x−2)(x+2)x+2f(x)=\frac{(x-2)(x+2)}{x+2}f(x)=x+2(x−2)(x+2)​

STEP 4

We can see that the factor (x+2)(x+2)(x+2) appears in both the numerator and the denominator. This means that the function is undefined at x=−2x=-2x=−2. However, because the factor is cancelled out, this point is a hole in the function, not a vertical asymptote.

STEP 5

To find the y-coordinate of the hole, we substitute x=−2x=-2x=−2 into the simplified function f(x)=x−2f(x)=x-2f(x)=x−2.f(−2)=−2−2f(-2)=-2-2f(−2)=−2−2

STEP 6

Calculate the y-coordinate of the hole.f(−2)=−2−2=−4f(-2)=-2-2=-4f(−2)=−2−2=−4So, the coordinates of the hole are (−2,−4)(-2, -4)(−2,−4).

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QuestionFind the holes in the rational function $f(x)=\frac{x^{2}-4}{x+2}$ . If none, state 'none'.

Assumptions1. The function is $f(x)=\frac{x^{}-4}{x+}$ . We are asked to find the coordinates of any holes in the function.

Now, we write the function in its factored form.
$f(x)=\frac{(x-2)(x+2)}{x+2}$

We can see that the factor $(x+2)$ appears in both the numerator and the denominator. This means that the function is undefined at $x=-2$ . However, because the factor is cancelled out, this point is a hole in the function, not a vertical asymptote.

To find the y-coordinate of the hole, we substitute $x=-2$ into the simplified function $f(x)=x-2$ .
$f(-2)=-2-2$

Calculate the y-coordinate of the hole.
$f(-2)=-2-2=-4$ So, the coordinates of the hole are $(-2, -4)$ .