What are the values of b for a minimum of x^2 + bx - 25?

[Cmillburg]

Find the values of b such that the function has the given minimum value.

$$ f(x) = x^2 + bx - 25$$;

Minimum value: - 34

 

[MarkFL]

Let's start with the fact that for a quadratic function in the form:

$$f(x)=ax^2+bx+c$$

The axis of symmetry (where the vertex, or global extremum occurs) is the line:

$$x=-\frac{b}{2a}$$

So, for the given function, where is the axis of symmetry? What is the value of the function for this value of $x$? What do you find when you equate this to -34?

 

[HallsofIvy]

Another way of doing it: complete the square.
x^2+ bx- 25
"x" is multiplied by "b". Half of that is b/2 and the square is b^2/4.
Add and subtract b^2/4:
x^2+ bx+ \frac{b^2}{4}- \frac{b^2}{4}- 25
(x+ \frac{b}{2})^2- (\frac{b^2}{4}+ 25

Since a square is never negative this will be minimum when the square is 0, that is, when x- b/2= 0 and that minimum will be -(\frac{b^2}{4}+ 25).

So solve -(\frac{b^2}{4}+ 25)= -34.