The values of h and k when f(x) = x^2 + 12x + 6 is in vertex form is -6 and -30
<h3>How to rewrite in vertex form?</h3>
The equation is given as:
f(x) = x^2 + 12x + 6
Rewrite as:
x^2 + 12x + 6 = 0
Subtract 6 from both sides
x^2 + 12x = -6
Take the coefficient of x
k = 12
Divide by 2
k/2 = 6
Square both sides
(k/2)^2 = 36
Add 36 to both sides of x^2 + 12x = -6
x^2 + 12x + 36= -6 + 36
Evaluate the sum
x^2 + 12x + 36= 30
Express as perfect square
(x + 6)^2 = 30
Subtract 30 from both sides
(x + 6)^2 -30 = 0
So, the equation f(x) = x^2 + 12x + 6 becomes
f(x) = (x + 6)^2 -30
A quadratic equation in vertex form is represented as:
f(x) = a(x - h)^2 + k
Where:
Vertex = (h,k)
By comparison, we have:
(h,k) = (-6,-30)
Hence, the values of h and k when f(x) = x^2 + 12x + 6 is in vertex form is -6 and -30
Read more about quadratic functions at:
brainly.com/question/1214333
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