By writing f(x) = 3x^2 - 15x + 11 in the form a(x+b)^2 + c, find the coordinates of the min turning point.

Check your answer by calculus.

Im really not sure about these questions.

Ive got as far as writing it as 3(x^2 - 5x) + 11. Is this in the right form? How can i go any further?

## Quadratic question Is my working right?

### Untouchable

Posted 01 December 2006 - 10:30 AM

By writing f(x) = 3x^2 - 15x + 11 in the form a(x+b)^2 + c, find the coordinates of the min turning point.

Check your answer by calculus.

Im really not sure about these questions.

Ive got as far as writing it as 3(x^2 - 5x) + 11. Is this in the right form? How can i go any further?

Writing f(x) = 3x² - 15x + 11 in the form a(x+b)² + c is the process of "completing the square".

This process can only work when the co-efficient (the number in front) of x² is 1.

Therefore re-aarange the function

f(x) = 3x² - 15x + 11

= 3(x² - 5x) + 11

Now try and "force" the square out of (x² - 5x) part.

= (x - 2.5)² - 2.5²

= (x - 2.5)² - 6.75

substitute this back into = 3(x² - 5x) + 11

= 3((x - 2.5)² - 6.25) + 11

= 3(x - 2.5)² - 18.75 + 11

= 3(x - 2.5)² - 7.75

f(x) = 3x² - 15x + 11 is now the form a(x+b)² + c

from this you can quickly get the turning point of the quadratic function