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

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To find the values of b for the quadratic function f(x) = x^2 + bx - 25 to have a minimum value of -34, the axis of symmetry is determined by x = -b/2. The minimum value occurs at this point and can be expressed as - (b^2/4 + 25). Setting this equal to -34 leads to the equation - (b^2/4 + 25) = -34. Solving this equation reveals the necessary values of b that achieve the specified minimum.
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Find the values of b such that the function has the given minimum value.

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

Minimum value: - 34
 
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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?
 
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.
 
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