core 1 maths help?

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?

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.

Reply 2

Chelsea12345

OP

12

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??

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.

Reply 4

username1753569

15

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.

Reply 5

TiernanW

15

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

(edited 7 years ago)

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

(x-a)^2+b

then I can see how much the parabola

x^2

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