What are the vertex and x-intercepts of the graph of y= x2 - 2x - 24? Select

Mathematics

1 answer:

guajiro [1.7K]3 years ago

4 0

Answer: x intercepts -4,0 6,0 vertex:1,-25

Step-by-step explanation:

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Find the horizontal asymptote

eimsori [14]

Answer:

Correct answer: h = 3/5 or y = 3/5

Step-by-step explanation:

Given:

h(x) = (3x² - 9x - 4) / (5x² - 4x + 8)

h = lim x -> ∞ (3x² - 9x - 4) / (5x² - 4x + 8) the numerator and denominator will be divided by x² and get

h = lim x -> ∞ (3 - 9(1/x) - 4 (1/x²)) / (5 - 4(1/x) + 8 (1/x²)

The terms 1/x and 1/x² when x strives ∞ they strives 0

x -> ∞ ⇒ 1/x -> 0 and x -> ∞ ⇒ 1/x² -> 0

h = (3 - 0 - 0) / (5 - 0 + 0)

Horizontal asymptote is h = 3/5 or y = 3/5

God is with you!!!

5 0

3 years ago

The ratio of the number of student in mr. Moore’s class compared to the number of student in miss Henry’s class is 5:4. The rati

Ahat [919]

Answer:

10:9

Step-by-step explanation:

First ratio is 5:4. 4 is equal to Henry class

If you múltiply the ratio by 2 you get

10:8, This ratio can be compared because now Henry class has 8

2(5:4)

10:8

8:9

Meaning that the answer is 10:9

7 0

3 years ago

If sec= 5/3 and the terminal point determined by is in quadrant 4, then

OLga [1]

$\bf sec(\theta)=\cfrac{1}{cos(\theta)}\\\\
-----------------------------\\\\
sec(\theta)=\cfrac{5}{3}\implies \cfrac{1}{cos(\theta)}=\cfrac{5}{3}\implies \cfrac{3}{5}=cos(\theta)
\\\\\\
\textit{now, }\theta\textit{ is in the 4th quadrant}$

so... notice the picture below, the angle in the fourth quadrant
now... notice, the cosine is just the distance the angle makes with the x-axis
so... using the pythagorean theorem, get the opposite side, from that triangle, keeping in mind that, "y" is negative in the 4th quadrant

once you get "y", get any other trigonometry value you need :)