# Why is (h,k) the 'Point of Inflection' and 'Centre' in Cubics and Circles? (1 Viewer)

#### _Anonymous

##### Member
As the title states, why is (h,k) the POI (Point of Inflection) and Centre in Cubics and Circles? I understand why they're there for Parabolas (have to find the maximum/minimum value), but what's the reasoning behind Circles and Cubics?

For a Cubic in y = a(x-h)^3 +k form, there are no maximum nor minimum values, so how/why does h,k mean it's the POI (Point of Inflection)? Similarly with Circles, (x-h)^2 + (y-k)^2 form the (h,k) are the 'centres' - why/how do they reflect the circles translating their centres (if that makes any sense)?

Say for example a question is like (x+2)^2 + (y-1)^2 = 4
(h,k) would be (-2,1) respectively. That would mean x and y have to be (-2,1), but wouldn't that lead to the equation being (0)^2 + (0)^2 = 4 - which is false? Same with Cubics y = a(x-3)^3 +k , the x would be 3 since it would lead to 3-3 = 0. Why do we make these values become a 0? I get it for Parabolas, but no clue why for Cubics or Circles.

I asked my teacher this and was told "It's just how it is" but I'm still curious as to why it's "how it is".

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#### Thatstudentm9

##### Member
why are u doing it 5.3 doesn't have this as far i know

#### Cena123

##### Member
Say for example a question is like (x+2)^2 + (y-1)^2 = 4
(h,k) would be (-2,1) respectively. That would mean x and y have to be (-2,1), but wouldn't that lead to the equation being (0)^2 + (0)^2 = 4 - which is false?
(-2,1) does not lie on the circle. It is the centre of the circle. Therefore LHS does not equal RHS when the points are subbed.

#### red152

##### Member
why are u doing it 5.3 doesn't have this as far i know
im in year 10 doing it too...

#### _Anonymous

##### Member
(-2,1) does not lie on the circle. It is the centre of the circle. Therefore LHS does not equal RHS when the points are subbed.
So when it's the centre, we imagine it being (0,0)? Why/how is it that (h,k) determine the centre in a circle?

#### jjHasm

##### Active Member
So when it's the centre, we imagine it being (0,0)? Why/how is it that (h,k) determine the centre in a circle?
It's good to see you are thinking over details like these. https://en.wikipedia.org/wiki/Circle <-- Click on that, scroll down "Equataions" then "Cartesian co-ordinates". It gives a solid explanation to your question. Try your best to understand. If you don't understand it, then don't dwell on it too much. If you do then kudos to you. Understanding stuff like this deeply, helps alot with later high school maths. It distinguishes you from the rest.

#### 1729

##### Active Member
So when it's the centre, we imagine it being (0,0)? Why/how is it that (h,k) determine the centre in a circle?
Because x^2 + y^2 = r^2 is centred at the origin. And to translate the centre h units to the right, we replace x with (x-h) and k units up, we replace y with (y-k), hence why the circle centred (h,k) has equation (x-h)^2 + (y-k)^2 = r^2

As for why we replace x with (x-h) for a horizontal rightward shift of h units, we are actually translating the coordinate system to the left which appears to be a rightward shift for the graph.

#### kawaiipotato

##### Well-Known Member
As the title states, why is (h,k) the POI (Point of Inflection) and Centre in Cubics and Circles? I understand why they're there for Parabolas (have to find the maximum/minimum value), but what's the reasoning behind Circles and Cubics?

For a Cubic in y = a(x-h)^3 +k form, there are no maximum nor minimum values, so how/why does h,k mean it's the POI (Point of Inflection)? Similarly with Circles, (x-h)^2 + (y-k)^2 form the (h,k) are the 'centres' - why/how do they reflect the circles translating their centres (if that makes any sense)?

Say for example a question is like (x+2)^2 + (y-1)^2 = 4
(h,k) would be (-2,1) respectively. That would mean x and y have to be (-2,1), but wouldn't that lead to the equation being (0)^2 + (0)^2 = 4 - which is false? Same with Cubics y = a(x-3)^3 +k , the x would be 3 since it would lead to 3-3 = 0. Why do we make these values become a 0? I get it for Parabolas, but no clue why for Cubics or Circles.

I asked my teacher this and was told "It's just how it is" but I'm still curious as to why it's "how it is".
$\bg_white \noindent (If I interpret your question correctly) A circle centred at (0,0) with radius r has cartesian equation x^2 + y^2 = r^2 (to see this: use Pythagoras' Theorem). Now let h,k be real numbers. \\ Also, for simplicity, let r=3 (so the equation we have is x^2 + y^2 = 3^2). \\ If you replace x with x-h in the equation, we get (x-h)^2 + y^2 = 3^2. This shifts the original circle that we had, to the right by h units.\\ To see this, take x=1 for example.\\ Then y = \pm \sqrt{8} (from the original equation).\\ Now focus on these two points (1,\sqrt{8}),(1,-\sqrt{8}) corresponding to x=1.\\ So let's take our new equation and make y the subject: y = \pm\sqrt{9-(x-h)^2}. \\ What happens if we plug our original x=1 value in? The y value would surely be different (test it: let h = 1 and x=1 (as we defined earlier)).$
$\bg_white \noindent To account for this change in y-value, we can do this by replacing x=1 with x=1+h (since if you plug this into the new y-value, we get y=\pm\sqrt{9-(1+h-h)^2} = \pm\sqrt{8}) As you can see, if we took our original equation, and replaced x with x-h, then for every y-value of the old equation, to match it with the new equation, we will need to increase our old x values by h, giving x+h. \\ Thus, this gives the geometrical interpretation of ''shifting'' the graph h units to the right. \\ As you can see, this will then shift the centre (0,0) to (h,0). \\ A similar argument is done for replacing y with y-k (which will then shift the new centre to (h,k)).$

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#### _Anonymous

##### Member
It's good to see you are thinking over details like these. https://en.wikipedia.org/wiki/Circle <-- Click on that, scroll down "Equataions" then "Cartesian co-ordinates". It gives a solid explanation to your question. Try your best to understand. If you don't understand it, then don't dwell on it too much. If you do then kudos to you. Understanding stuff like this deeply, helps alot with later high school maths. It distinguishes you from the rest.
Thanks for the link, think I've got it now. I then found a similar solution to the Circle problem in a MX1 book which helped as well. Do you know any websites explaining for as to why (h,k) is the POI for Cubics?

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#### _Anonymous

##### Member
Because x^2 + y^2 = r^2 is centred at the origin. And to translate the centre h units to the right, we replace x with (x-h) and k units up, we replace y with (y-k), hence why the circle centred (h,k) has equation (x-h)^2 + (y-k)^2 = r^2

As for why we replace x with (x-h) for a horizontal rightward shift of h units, we are actually translating the coordinate system to the left which appears to be a rightward shift for the graph.
Thanks for your help. Do you know why (h,k) is the Point of Inflection for Cubics though?

#### InteGrand

##### Well-Known Member
Thanks for your help. Do you know why (h,k) is the Point of Inflection for Cubics though?
Because the origin is a point of inflection for the un-translated cubic (y = x3).

(Similarly, the origin is the centre of the un-translated circle (x2 + y2 = a2), so (h, k) is the centre of the translated circle (x-h)2 + (y-k)2 = a2. In other words, all you're doing is translating things; the "nature" of the corresponding points are the same.)

#### kawaiipotato

##### Well-Known Member
Thanks for your help. Do you know why (h,k) is the Point of Inflection for Cubics though?
$\bg_white \noindent Following from what InteGrand said (and from what I said earlier as well),
in general if you have a function in terms of x,f(x), then f(x-a) is the graph of f(x) shifted to the right a units.
So take a cubic function f(x)=x^3. This has POI at (0,0). \\Then g(x) := f(x-h) = (x-h)^3 means we shift the graph to the right by h units. \\So what's the new y-value? It's still 0 since if we let x=h, then g(h) = (h-h)^3 = 0.\\ So the new POI is (h,0). \\If we add a constant k to the equation to give p(x) = (x-h)^3 + k, then simply p(h) = (h-h)^3 + k = k. The new POI is (h,k).$