Consider the equation below. y=3x^2+30x+71 use completing the square to rewrite the given equation and reveal the extreme value y=3(x+blank)^2+blank the extreme value of the equation is at (blank, blank)

Equation in square form:

Extreme value:

Step-by-step explanation:

We are given

we can complete square

we can use formula

now, we can add and subtract 5^2

So, we get equation as

Extreme values:

we know that this parabola

and vertex of parabola always at extreme values

so, we can compare it with

where

vertex=(h,k)

now, we can compare and find h and k

we get

h=-5

k=-4

so, extreme value of this equation is

Top row: reveal extreme value 5 , -4

Bottom row: extreme value -5, -4

Step-by-step explanation:

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation:

provide more details to solve it

step-by-step explanation: