Find the numbers b and c in the quadratic equation
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What is the condition for max/min of a second order function (parabola)?
@Subhotosh Khan: I think everything is written down and I got to the answer without asking about anything. But of course, no offence, I understand some people are at a lower level than me 
@Bob90:
You know that your minimum y=3 when x=4, so p=4, q=3. You have a vertex P(p,q), when p and q are the coordinates (x,y) of a vertex.
You can then write that your f(x)=x2+bx+c=(x-p)2+q=(x-4)2+3.
Then you get f(x)=x2-8x+16+3 , so you can now read from the equation that b=-8 and c=19.
PS: I'm not a native speaker so if anything is unclear or incorrect written - please write, I'm still improving my English.
@Bob90:
You know that your minimum y=3 when x=4, so p=4, q=3. You have a vertex P(p,q), when p and q are the coordinates (x,y) of a vertex.
You can then write that your f(x)=x2+bx+c=(x-p)2+q=(x-4)2+3.
Then you get f(x)=x2-8x+16+3 , so you can now read from the equation that b=-8 and c=19.
PS: I'm not a native speaker so if anything is unclear or incorrect written - please write, I'm still improving my English.
The question i have in the book is literally :
"Find the numbers b and c in the equation of the function f : y = x2 + bx + c so that f reaches its
minimal value 3 at x = 4."
I know to find the vertex it is : -b/2a...
But to find the answer to this question i need help on how to even start it.
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@Subhotosh Khan: I think everything is written down and I got to the answer without asking about anything. But of course, no offence, I understand some people are at a lower level than me
@Bob90:
You know that your minimum y=3 when x=4, so p=4, q=3. You have a vertex P(p,q), when p and q are the coordinates (x,y) of a vertex.
You can then write that your f(x)=x2+bx+c=(x-p)2+q=(x-4)2+3.
Then you get f(x)=x2-8x+16+3 , so you can now read from the equation that b=-8 and c=19.
PS: I'm not a native speaker so if anything is unclear or incorrect written - please write, I'm still improving my English.
brejalis:
I know everything is given and the answer can be derived from the given statements. I am at a pretty high level - I have a Ph.D in engineering. How about you?
However, I wanted the OP to "discover" the process to the answer - not spoon-feed.
D
Deleted member 4993
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You can transform this equation to an equation of parabola by "completing the square".
y = (x + b/2)^2 - (b^2/4 -c)
Where is the minimal value of the equation above?
@Subhotosh Khan: I think everything is written down and I got to the answer without asking about anything. But of course, no offence, I understand some people are at a lower level than me
@Bob90:
You know that your minimum y=3 when x=4, so p=4, q=3. You have a vertex P(p,q), when p and q are the coordinates (x,y) of a vertex.
You can then write that your f(x)=x2+bx+c=(x-p)2+q=(x-4)2+3.
Then you get f(x)=x2-8x+16+3 , so you can now read from the equation that b=-8 and c=19.
PS: I'm not a native speaker so if anything is unclear or incorrect written - please write, I'm still improving my English.
Thank you very much to you both, i could solve all the other questions of the chapter in my textbook but this one was tricky...I will work on the question some more to make sure i understand it properly.