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

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?

- Jan 29, 2012

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[tex]x^2+ bx- 25[/tex]

"x" is multiplied by "b". Half of that is b/2 and the square is [tex]b^2/4[/tex].

Add and subtract [tex]b^2/4[/tex]:

[tex]x^2+ bx+ \frac{b^2}{4}- \frac{b^2}{4}- 25[/tex]

[tex](x+ \frac{b}{2})^2- (\frac{b^2}{4}+ 25[/tex]

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 [tex]-(\frac{b^2}{4}+ 25)[/tex].

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