How do I find the vertex of a parabola?

Answer 1

If you have it in vertex form it is really easy.

Vertex form is:
y = a(x - h)² + k

In this form, the vertex is at the point (h, k).

For example, let's say your equation was:
y = x² + 4

This is equivalent to:
y = 1(x - 0)² + 4

So you should be able to readily see that the vertex is (0, 4).

Similarly if you had:
y = -2x² + 9

This is the same as:
y = -2(x - 0)² + 9

So the vertex is (0, 9)

Now what if the equation isn't in vertex form? Say it was:
y = x² + 6x + 2

You would need to complete the square to get it into vertex form.

Start by moving the 2 to the left side:
y - 2 = x² + 6x

Now if there was a coefficient on the x² term you would divide both sides by that, but there isn't so you can skip that step.

Take the coefficient on the x term (6), take half (3) and square it (9). This is the number you need to add to both sides.
y - 2 + 9 = x² + 6x + 9
y + 7 = x² + 6x + 9

Write the right side as a perfect square:
y + 7 = (x + 3)²

Finally move the 7 back to the right and you have it in vertex form:
y = (x + 3)² - 7.

Comparing to:
y = a(x - h)² + k

a = 1
h = -3 (remember x + 3 = x - (-3))
k = -7

So here the vertex is (h, k) = (-3, -7)

Answer 2

Let f(x) = ax^2+bx+c, then the x-value of the vertex is x= - b/(2a).

Once you have this x-value, plug it into f(x) to get the corresponding y-value of the vertex.

For example:

f(x) = 3x^2+12x-2 then x= -12/(2*3) = -12/6 = -2

f(-2) = 3*(-2)^2+12*2-2= 3*4+24-2=12+22=34

so the vertex is (-2, 34).

If you have the equation in standard form f(x) = a(x-b)^2 +c

Then (b,c) is the vertex.

Examples: f(x) = 2(x-3)^2+4 has vertex (3,4).

f(x) = 5(x+7)^2-9 has vertex (-7,-9).

Notice that for x-3, the x-value of the vertex is 3 and for x+7 it's -7. These change sign, but the 4 and -9 for the y-value stay the same sign.

Answer 3

The first answer you got for this is only half correct. You get the x point of the vertex from -b/2a. In order to get the y point, you need to plug the x value back into the orginial problem.

Example: x^2-2x+3
-b would be 2 and 2a would be 2 also. So, the x value would be 2/2=1.
Put that value into the original and you have
1^2-2(1)+3=2
The vertex of this parabola is (1,2)

Answer 4

A parabola is a symmetric curve.

The intersection of the line of symmetry with the parabola is the vertex.

Answer 5

Take the derivative.
set it equal to zero
solve for x
substitute that value for x in the original equation
solve for y

Answer 6

put the equation in the form y-k=a(x-h)^2
and the vertex will begiven by (h,k)

Answer 7

-b / 2a