H=48t-16t squared 32=48t-16t sqaured
1 answer:
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Answer:
The horizontal asymptote can be described by the line y = 6
The vertical asymptote can be described by the line x = -2
Step-by-step explanation:
* <em>Lets the meaning of vertical and horizontal asymptotes</em>
- <u><em>Vertical asymptotes</em></u> are vertical lines which correspond to the zeroes
of the denominator of a rational function
- <u><em>A horizontal asymptote</em></u> is a y-value on a graph which a function
approaches but does not actually reach
- If the degree of the numerator is less than the degree of the
denominator, then there is a horizontal asymptote at y = 0
- If the degree of the numerator is greater than the degree of the
denominator, then there is no horizontal asymptote
- If the degree of the numerator is equal the degree of the denominator,
then there is a horizontal asymptote at y = leading coefficient of the
numerator ÷ leading coefficient of the denominator
* <em>Lets solve the problem</em>
∵
∵ The numerator is 6x
∵ The denominator is x + 2
∴ The numerator and the denominator have same degree
∵ The leading coefficient of the numerator is 6
∵ The leading coefficient of the denominator is 1
∴ There is a horizontal asymptote at y = 6/1
∴ <em>The horizontal asymptote can be described by the line y = 6</em>
- Put the denominator equal zero to find its zeroes
∵ The denominator is x + 2
∴ x + 2 = 0
- Subtract 2 from both sides
∴ x = -2
∴ <em>The vertical asymptote can be described by the line x = -2</em>