# Turning Point of a Parabola

• Nov 8th 2008, 05:35 PM
J4553
Turning Point of a Parabola
Hello,

How do I find the turning point of a parabola?

• Nov 8th 2008, 06:05 PM
mr fantastic
Quote:

Originally Posted by J4553
Hello,

How do I find the turning point of a parabola?

Complete the square.

Alternatively, use calculus.

Post a more specific question (and state your mathematical background).
• Nov 8th 2008, 06:09 PM
Shyam
Quote:

Originally Posted by J4553
Hello,

How do I find the turning point of a parabola?

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

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

where (h, k) is vertex.

If, suppose equation of a parabola is

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

Then, complete the square, so, eqn becomes,

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

so, vertex = (1, 2)

so, turning point = (1, 2)

Did you get it now???
• Nov 8th 2008, 06:17 PM
J4553
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 PM
euclid2
Quote:

Originally Posted by J4553
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?

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 PM
J4553
Perfect - thanks!
• Nov 8th 2008, 06:33 PM
Shyam
Quote:

Originally Posted by J4553
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?

$y = 3x^2-6x+5$
$x = \frac{-b}{2a}=\frac{-(-6)}{(2)(3)}=1$
$y = 3(1)^2-6(1)+5$