Thread: Turning Point of a Parabola

1. Turning Point of a Parabola

Hello,

How do I find the turning point of a parabola?

2. 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).

3. 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???

4. 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?

5. 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

6. Perfect - thanks!

7. 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?

If, suppose equation of a parabola is

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

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

$y = 3(1)^2-6(1)+5$

y = 2

so, you have (1, 2) as the vertex or turning point of parabola