A question for C1

I get where they got y=x^2 + 2x - 5 from . bit i dont get is (x+1)^2 - 6 . minimum at (-1,-6). how do they get that

thanks guys

You need to learn how to complete the square.

If a quadratic is in the form $y = k(x-a)^2 + b$, then the coordinates of the vertex are (a, b).

Why can't you factorise?

Try it. It doesn't work. (with integer values anyway)
Completing the square in situations where factorising isn't possible is easier. Also you get the minimum point, where as if you factorised, you'd only get the roots of the equation.

Once you have completed the square, how do you use it to get the minimum point?

-Thanks for your reply
- Yes you are right it doesn't work.
- Once you have completed the square, how do you use it to get the minimum point?
See here: http://www.thestudentroom.co.uk/showthread.php?p=28857201#post28857201
- Also if I did factorise it I would have put x=0 and found the y intercept and put y=0 to find the x -axis intercepts . So basically I would have found were the curve is crossing the axes, so why in this case do you use the minimum point instead or using the intercepts to sketch the curve? And can you even use the minimum point to sketch the curve?
Yes, the y-intercept is when x=0.
Yes, if you factorised, you are essentially putting y=0 into the equation.
In this case, you're using the minimum point instead as you don't know the x-intercepts. You can use the minimum point to sketch a curve, as you should know the general shape, where the minimum point lies, and the y-intercept, so you can sorta guage whereabouts the curve will cross the x-axis.
- Why havent they labelled on the minimum point on their graph?
The graph was drawn in part a, not part d. You worked out the minimum point in part d, not part a, so it wasn't labelled in the diagram. If you do know the minimum point, its common practice to label it.

My post in bold.

Thanks for the reply. very helpfull

when i complete the square for y=x^2 + 2x - 5 i get = (x+1)^2 - 1 ?

You need to learn how to complete the square.

If a quadratic is in the form $y = k(x-a)^2 + b$, then the coordinates of the vertex are (a, b).

Right, so I completed the square now.

= (x+1)^2 -6

so the minmum point should be (1, -6) but its (-1,-6) , why is it -1 and not +1 after all inside the bracket it says +1