Need the awnser to this question awnser

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

Extreme values of the given equation y=3x^2+30x+71 are at (-5,-4)

Step-by-step explanation:

We need to complete the square and find extreme values of the given equation:

y=3x^2+30x+71

First, completing the square:

Taking 3 common from first 2 terms:

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

We need to make the form: a^2+2ab+b^2 for completing square:

Since We have taking 3 common, we need to be adding and subtracting 3(5)^2 for completing square.

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

And we know that a^2+2ab+b^2 = (a+b)^2

y = 3(x+5)^2 +71-3(25)

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

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

Looking at the equation, this represents a parabola

The general equation of parabola is:

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

The vertex of parabola is at (h,k) and the vertex of parabola is always at extreme values. So, finding the vertex of the equation y = 3(x+5)^2 - 4

Comparing the standard equation of parabola with y = 3(x+5)^2 - 4

so, h = -5 and k = -4

So, Extreme values of the given equation y=3x^2+30x+71 are at (-5,-4)

1 3 5

step-by-step explanation:

i took the test: )