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docker41 [41]
2 years ago
9

What is the equation of the function represented by this graph?

Mathematics
1 answer:
gayaneshka [121]2 years ago
6 0

Answer:

\displaystyle g(x) = -(x + 2)(x - 4)

Step-by-step explanation:

First, look at the graph to determine whether it opens down or up, which in this case is <em>down</em>, so you insert a negative in front of the expression. Next, locate your <em>x-intercepts</em> [<em>zeros</em> (where the graph insects the x-axis)]. You will see that the graph intersects \displaystyle [-2, 0] and \displaystyle [4, 0], so what you do is insert the zeros' OPPOCITES into the equation because according to the vertex equation, \displaystyle y = A[x - h]^2 + k, that negative in front of \displaystyle h gives you the OPPOCITE outcome, so be careful.

I am joyous to assist you at any time.

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Answer:

Step-by-step explanation:

Limit refers to the value that the function approaches as the input approaches some value.

We say \displaystyle \lim_{x\rightarrow a}f(x)=L, if f(x) approaches L as x approaches 'a'.

(a)

\displaystyle \lim_{x\rightarrow 5}\left ( \frac{f(x)-8}{x-5} \right )=4\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-\displaystyle \lim_{x\rightarrow 5}8}{\displaystyle \lim_{x\rightarrow 5}x-\displaystyle \lim_{x\rightarrow 5}5}=4\\

\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-8}{\displaystyle \lim_{x\rightarrow 5}x-5}=4\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4\left ( \displaystyle \lim_{x\rightarrow 5}x-5 \right )\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4\displaystyle \lim_{x\rightarrow 5}x-4(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4(5)-4(5)\\

\displaystyle \lim_{x\rightarrow 5}f(x)-8=20-20=0\\\displaystyle \lim_{x\rightarrow 5}f(x)=8

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\displaystyle \lim_{x\rightarrow 5}\left ( \frac{f(x)-8}{x-5} \right )=7\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-\displaystyle \lim_{x\rightarrow 5}8}{\displaystyle \lim_{x\rightarrow 5}x-\displaystyle \lim_{x\rightarrow 5}5}=7\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-8}{\displaystyle \lim_{x\rightarrow 5}x-5}=7\\

\displaystyle \lim_{x\rightarrow 5}f(x)-8=7\left ( \displaystyle \lim_{x\rightarrow 5}x-5 \right )\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=7\displaystyle \lim_{x\rightarrow 5}x-7(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=7(5)-7(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=35-35=0\\\displaystyle \lim_{x\rightarrow 5}f(x)=8

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Answer:

1. it is not a direct variation because the value <em>k</em> is not kept throughout the ordered pairs

2. -a^5 - 2a³ + 2a², leading coefficient is -1

Step-by-step explanation:

1. to find the direct variation, we will use y/x = k with k representing our constant throughout the chart. if every value is equal to k, it is a direct variation

for this problem we will assume that the cup sizes are x and the cost is y

so the ordered pairs are (8, 0.90), (12, 1.35), (16, 1.80), and (24, 2.25)

.90/8 = 0.1125 <-- this is our <em>k</em> value for which every result has to be equal to

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since the value <em>k</em> was not kept throughout the ordered pairs, this is not a direct variation

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we have the following polynomial: 2a² - a^5 -2a³

to write it in standard form, we will go in descending order of the exponents. the largest exponent being first

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the leading coefficient of the polynomial is -1, as there is a negative sign in front of the newly formed polynomial

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