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# core 1 maths help?

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1. I need to find the co-ordinate of the stationary point of the curve y=x^2-4x+5.
so to find the x co-ordinates do i factorise or complete the square? i tried factorisng the equation but it doesnt work?
2. (Original post by Chelsea12345)
I need to find the co-ordinate of the stationary point of the curve y=x^2-4x+5.
so to find the x co-ordinates do i factorise or complete the square? i tried factorisng the equation but it doesnt work?
Well you can't always factorise. With quadratics and their stationary points you have 2 options:

1) Complete the square to find the x-coordinate and then by inspection (or substitution) find the y-coordinate

2) Differentiate the function and set the differential equal to 0 and solve it. Setting it to 0 will ensure that the rate of change is 0 as to be expected at a stationary point.
3. (Original post by RDKGames)
Well you can't always factorise. With quadratics and their stationary points you have 2 options:

1) Complete the square to find the x-coordinate and then by inspection (or substitution) find the y-coordinate

2) Differentiate the function and set the differential equal to 0 and solve it. Setting it to 0 will ensure that the rate of change is 0 as to be expected at a stationary point.
if you complete the square you only get one x-cordinate,dont you need 2 of them??
4. (Original post by Chelsea12345)
if you complete the square you only get one x-cordinate,dont you need 2 of them??
Yeah, you get the same through differentiation. You find the x-coordinate first and then plug it back through the quadratic to see what the y-coordinate is.
5. (Original post by Chelsea12345)
if you complete the square you only get one x-cordinate,dont you need 2 of them??
Hi, as this is a quadratic, your only stationary point is your vertex. Complete the square to find this.
6. (Original post by RDKGames)
Yeah, you get the same through differentiation. You find the x-coordinate first and then plug it back through the quadratic to see what the y-coordinate is.
EDIT: Never mind, I was being an idiot LOLOL
7. (Original post by TiernanW)
When you complete the square isn't it just the equation re-wrote in another form? So aren't you then finding the points where it crosses the x-axis or am I being an idiot?
Completing the square is indeed a different form to write a quadratic, and no it does not necessarily mean you are finding the roots of the equation.

If I put some arbitrary quadratic in the form of then I can see how much the parabola has been shifted along the x-axis (in this case it has been shifted by vector [a,0] as far as the horizontal transformation is concerned). Since the x-coordinate of the stationary point of is at 0, the shift will be the same for it to get the stationary point of .

If you want to find the roots, you would have to solve
8. (Original post by slowdive)
Hi, as this is a quadratic, your only stationary point is your vertex. Complete the square to find this.
okay thankyou!!

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