IB Maths HL Completing the square

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IB Maths HL Completing the square

IB Mathematics HL– Complete the square

How can we complete the square and find the vertex of the following quadratic function:

$f(x)=2x^2-9x+15$

Thanks
Max

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Joined: Mon Jan 28, 2013 8:06 pm

Re: IB Maths HL Completing the square

IB Mathematics HL – Complete the square

Completing the square is a procedure for reformatting quadratic functions.
Quadratic Functions of the form $f(x)=ax^2+bx+c$ can be rewritten as

$f(x)=a(x-h)^2+k$
The major reason for doing this is that it allows us to easily see the vertex of the curve $f(x)=ax^2+bx+c$ which is located at the point $(h,k)$.
The general method of completing the square can be shown like this:

$f(x)=ax^2+bx+c=a(x^2+\frac{b}{a}x)+c=$

$=a((x+\frac{b}{2a})^2-(\frac{b}{2a})^2)+c=$

$=a(x-(-\frac{b}{2a}))^2-\frac{b^2}{4a}+c$

Therefore the x-coordinate of the vertex is $h=-\frac{b}{2a}$

$f(x)=2x^2-9x+15=2(x^2+\frac{-9}{2}x)+15=$

$=2((x+\frac{-9}{2(2)})^2-(\frac{-9}{2(2)})^2)+15=$

$=2(x-(\frac{9}{2(2)}))^2-\frac{(-9)^2}{8}+15=$

$=2(x-(\frac{9}{4}))^2-\frac{81}{8}+15=$

$=2(x-(\frac{9}{4}))^2-\frac{81}{8}+\frac{15(8)}{8}=$

$=2(x-(\frac{9}{4}))^2-\frac{81}{8}+\frac{120}{8}=$

$=2(x-(\frac{9}{4}))^2+\frac{39}{8}=$

Therefore the coordinates of the vertex are $( \frac{9}{4}, \frac{39}{8})$

Hope these help!!
miranda

Posts: 268
Joined: Mon Jan 28, 2013 8:03 pm

Re: IB Math HL Completing the square

Thank you miranda!!
Max

Posts: 0
Joined: Mon Jan 28, 2013 8:06 pm