the minimum point on a parabola is its vertex. So, find the vertex.
For y = ax^2+bx+c, the vertex is at x = -b/2a. In this case, that is
x = -12/2 = -6
AT x=-6, y = -31
This can be seen by rearranging the equation into the vertex form.
x^2+12x+5
= (x^2+12x+36) + 5 - 36
= (x+6)^2 - 31
What is the minimum possible value for y=x^2+12x+5? I am not sure where to even start. Thank you!
2 answers
Calculus is amazing. Take the derivative of the quadratic and get 2x+12. Set that equal to 0 to get x=-6. Plug that back in to the original equation to get y=-31
1 answer