Hello,

How do I find the turning point of a parabola?

Thanks in advance

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- Nov 8th 2008, 05:35 PMJ4553Turning Point of a Parabola
Hello,

How do I find the turning point of a parabola?

Thanks in advance - Nov 8th 2008, 06:05 PMmr fantastic
- Nov 8th 2008, 06:09 PMShyam
The turning point of a parabola is the

**vertex**of the parabola.

For the given equation of parabola, you can find the vertex by completing the square in the form

**$\displaystyle y = a(x-h)^2+k$**

where**(h, k)**is**vertex**.

If, suppose equation of a parabola is

$\displaystyle y = 3x^2-6x+5$

Then, complete the square, so, eqn becomes,

$\displaystyle y = 3(x-1)^2+2$

so, vertex = (1, 2)

so, turning point = (1, 2)

Did you get it now??? - Nov 8th 2008, 06:17 PMJ4553
Shyam - no, I'm afraid that just confuses me even more.

I think my lecturer gave the formula: -b/2a to find the x-component of the vertex, then may have said to substitute that value into another equation to find the y-component.

Does this sound right or am I way off? - Nov 8th 2008, 06:23 PMeuclid2
that formula is correct, for example, take this function;

y=x^2+6x+8

b=6, a=1

-b/2a=-6/2(1)=-3

sub -3 into the function to get the value of y

y=(-3)^2+6(-3)+8=-1

therefore, the vertex for this example is (-3,-1)

Shyam is simply using another method then the one you are instructed to use - Nov 8th 2008, 06:30 PMJ4553
Perfect - thanks!

- Nov 8th 2008, 06:33 PMShyam