Answer : d. minimum (-1,3)
\(f(x) = 5x^2 + 10x + 8\)
The vertex form of quadratic function is
\(f(x) = a(x-h)^2 + k\), where (h,k) is the vertex
To get vertex form we apply completing the square method
To apply completing the square method , there should be only x^2
So we factor out 5 from from first two terms
\(f(x) = 5(x^2 + 2x) + 8\)
Now we take the number before x (coefficient of x) and divide by 2
\(\frac{2}{2}\) =1
Now square it
\((1)^2 =1\)
Add and subtract 1 inside the parenthesis
\(f(x) = 5(x^2 + 2x + 1 - 1) + 8\)
Now we take out -1 by multiplying 5
\(f(x) = 5(x^2 + 2x + 1) -5 + 8\)
\(f(x) = 5(x^2 + 2x + 1) + 3\)
Now we factor x^2 +2x+1 as (x+1)(x+1)
\(f(x) = 5(x+1)(x+1) + 3\)
\(f(x) = 5(x+1)^2 + 3\)
h=-1 and k=3
So vertex is (-1,3)
When the value of 'a' is negative , then it is a maximum
When the value of 'a' is positive , then it is a minimum
\(f(x) = 5x^2 + 10x + 8\) is in the form of \(f(x) = ax^2 + bx + c\)
The value of a is 5
5 is positive so it is a minimum
f(x) is minimum at point (-1,3)
