QuestionFind the domain of $H(x)=\frac{5 x-30}{36-x^{2}}$ and its vertical asymptote(s). Choices included.

Studdy Solution

STEP 1

Assumptions1. The function is given by $H(x)=\frac{5x-30}{36-x^{}}$ . We are asked to find the domain of the function, which is the set of all possible x-values.
3. We are also asked to find the vertical asymptote(s) of the function, which are the x-values where the function approaches infinity or negative infinity.

STEP 2

The domain of a function is the set of all possible x-values that will not make the function undefined. A function becomes undefined when the denominator of a fraction is zero because division by zero is undefined.
So, we need to find the x-values that will make the denominator of the function zero.
Set the denominator equal to zero and solve for x $36-x^{2} =0$

STEP 3

This is a quadratic equation in the form of $ax^{2} + bx + c =0$ . We can solve it by factoring, completing the square, or using the quadratic formula. In this case, factoring is the easiest method.
Factor the equation $(6-x)(6+x) =0$

STEP 4

Setting each factor equal to zero gives the solutions to the equation $6-x =0$
and $6+x =0$

STEP 5

olving these equations gives $x =$
and $x = -$

STEP 6

These are the x-values that make the denominator zero, so they are excluded from the domain of the function. Therefore, the domain of the function is all real numbers except -6 and6. This corresponds to choice A in the problem statement $\{x \mid x \neq-6,6\}$

STEP 7

Next, we need to find the vertical asymptote(s) of the function. A vertical asymptote is a vertical line at an x-value where the function approaches infinity or negative infinity. This happens at the x-values that make the denominator zero, which we already found in the previous steps.
Therefore, the vertical asymptotes of the function are $x = -6$ and $x =6$ . This corresponds to choice A in the problem statementThe vertical asymptote(s) is(are) $x = -6$ and $x =6$ .

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QuestionFind the domain of $H(x)=\frac{5 x-30}{36-x^{2}}$ and its vertical asymptote(s). Choices included.

Studdy Solution

STEP 1

STEP 2

STEP 3

This is a quadratic equation in the form of $ax^{2} + bx + c =0$ . We can solve it by factoring, completing the square, or using the quadratic formula. In this case, factoring is the easiest method.
Factor the equation $(6-x)(6+x) =0$

STEP 4

Setting each factor equal to zero gives the solutions to the equation $6-x =0$
and $6+x =0$

STEP 5

olving these equations gives $x =$
and $x = -$

STEP 6

These are the x-values that make the denominator zero, so they are excluded from the domain of the function. Therefore, the domain of the function is all real numbers except -6 and6. This corresponds to choice A in the problem statement $\{x \mid x \neq-6,6\}$

STEP 7

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QuestionFind the domain of H(x)=5x−3036−x2H(x)=\frac{5 x-30}{36-x^{2}}H(x)=36−x25x−30​ and its vertical asymptote(s). Choices included.

Studdy Solution

STEP 1

STEP 2

STEP 3

This is a quadratic equation in the form of ax2+bx+c=0ax^{2} + bx + c =0ax2+bx+c=0. We can solve it by factoring, completing the square, or using the quadratic formula. In this case, factoring is the easiest method.Factor the equation(6−x)(6+x)=0(6-x)(6+x) =0(6−x)(6+x)=0

STEP 4

Setting each factor equal to zero gives the solutions to the equation6−x=06-x =06−x=0and6+x=06+x =06+x=0

STEP 5

olving these equations givesx=x =x=andx=−x = -x=−

STEP 6

These are the x-values that make the denominator zero, so they are excluded from the domain of the function. Therefore, the domain of the function is all real numbers except -6 and6. This corresponds to choice A in the problem statement{x∣x≠−6,6}\{x \mid x \neq-6,6\}{x∣x=−6,6}

STEP 7

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionFind the domain of $H(x)=\frac{5 x-30}{36-x^{2}}$ and its vertical asymptote(s). Choices included.

This is a quadratic equation in the form of $ax^{2} + bx + c =0$ . We can solve it by factoring, completing the square, or using the quadratic formula. In this case, factoring is the easiest method.
Factor the equation $(6-x)(6+x) =0$

Setting each factor equal to zero gives the solutions to the equation $6-x =0$
and $6+x =0$

olving these equations gives $x =$
and $x = -$

These are the x-values that make the denominator zero, so they are excluded from the domain of the function. Therefore, the domain of the function is all real numbers except -6 and6. This corresponds to choice A in the problem statement $\{x \mid x \neq-6,6\}$