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NeX [460]
3 years ago
10

Find the minimum or maximum value of the function g(x)= -(x-2)^2-3

Mathematics
1 answer:
Nady [450]3 years ago
3 0
Neato

assuming you mean
g(x)=-1(x-2)^2-3
remember
in form
f(x)=a(x-h)^2+k
a is leading coficient
(h,k) is vertex
k will be the minimum or max value of the function
if a is positive, then the parabola opens up and k is minimum
if a is negative, then the parabola opens down and k is max

given
g(x)=-1(x-2)^2-3
a is negative
the parabola opens down and k is the max value
it is -3
the max value is -3, which occors at x=2
g(2)=-3 is max
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3 years ago
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Answer:

see explanation

Step-by-step explanation:

let pq = x

given oq - pq = 1 then oq = 1 + x

Using Pythagoras' identity, then

(oq)² = 7² + x²

(1 + x)² = 49 + x² ( expand left side )

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2x = 48 ( divide both sides by 2 )

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and oq = 1 + x = 1 + 24 = 25 ← hypotenuse

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cosq = \frac{adjacent}{hypotenuse} = \frac{24}{25}



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3 years ago
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\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

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