Answer:
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)
