What is the end behavior of f(x)=4x^4-3x^2?

Mathematics

1 answer:

galben [10]3 years ago

4 0

Answer:

as x→∞, y→∞

as x→-∞, y→∞

Step-by-step explanation:

To tell what the end-behavior is, look at the x with highest exponent value in the function.

f(x) = 4x⁴ - 3x²

The highest power is x⁴.

Our highest exponent value is 4.

If the highest power is an even number, the end behavior is:

as x→∞, y→∞

as x→-∞, y→∞

this is always true. It's part of the rule for end-behavior.

The end behavior graph for an even power will look like the graph for x².

This is also always true. It will always look like this if the highest power is an even number.

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sarah used 3/4 pound of blueberries to make 1/2 cup of jam. how many pounds of blueberries would she need to make a cup of jam?

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Since we have given that

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As we need to find the number of pounds of blueberries she would need to make a cup of jam.

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3 years ago

In a free-fall experiment, an object is dropped from a height of h = 400 feet. A camera on the ground 500 ft from the point of i

PilotLPTM [1.2K]

Hmmm the object, is at rest, when dropped, so it has a velocity of 0 ft/s

the only force acting on the object, is gravity, using feet will then be -32ft/s²,

was wondering myself on -32 or 32.. but anyhow... we'll settle for the negative value, since it seems to be just a bit of convention issues

so, we'll do the integral to get v(t) then

$\bf \displaystyle \int -32\cdot dt\implies -32t+C
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\textit{object moves from \underline{rest}, so velocity is 0 at 0secs}
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-32(0)+C=0\implies C=0\implies \boxed{v(t)=-32t}
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\textit{now to get the positional s(t)}
\\\\\\
\displaystyle \int -32t\cdot dt\implies -16t^2+C
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\textit{the initial \underline{position} was 400ft away at 0secs}
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-16(0)^2+C=400\implies C=400\implies \boxed{s(t)=-16t^2+400}$

when will it reach the ground level? let's set s(t) = 0

$\bf s(t)=-16t^2+400\implies 0=-16t^2+400\implies \cfrac{-400}{-16}=t^2
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25=t^2\implies \boxed{5=t}$

part B) check the picture below