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Kisachek [45]
3 years ago
5

Consider the equation below.

Mathematics
2 answers:
Korvikt [17]3 years ago
6 0

Answer:

Equation in square form:

y=3(x+5)^2-4

Extreme value:

(h,k)=(-5,-4)

Step-by-step explanation:

We are given

y=3x^2+30x+71

we can complete square

y=3(x^2+10x)+71

we can use formula

a^2+2ab+b^2=(a+b)^2

y=3(x^2+2\times x\times 5)+71

now, we can add and subtract 5^2

y=3(x^2+2\times x\times 5+5^2-5^2)+71

y=3(x^2+2\times x\times 5+5^2)-3\times 5^2+71

y=3(x+5)^2-75+71

So, we get equation as

y=3(x+5)^2-4

Extreme values:

we know that this parabola

and vertex of parabola always at extreme values

so, we can compare it with

y=a(x-h)^2+k

where

vertex=(h,k)

now, we can compare and find h and k

y=3(x+5)^2-4

we get

h=-5

k=-4

so, extreme value of this equation is

(h,k)=(-5,-4)

choli [55]3 years ago
4 0

Answer:

Top row: reveal extreme value 5 , -4

Bottom row: extreme value -5, -4

Step-by-step explanation:

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