## IB Maths HL Completing the square

Discussions for the Core part of the syllabus. Algebra, Functions and equations, Circular functions and trigonometry, Vectors, Statistics and probability, Calculus. IB Maths HL Revision Notes

### 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

Posts: 0
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