# Finding B when given minimum

#### Cmillburg

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

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

Minimum value: - 34

#### MarkFL

Staff member

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

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

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

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

##### Well-known member
MHB Math Helper
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 $$-(\frac{b^2}{4}+ 25)= -34$$.