Vertex of a Parabola| Formula | Solved Examples

The vertex of a parabola is the point where the parabola intersects its axis of symmetry; this point represents its maximum or minimum value. It is the highest or lowest point on the graph, depending on whether the parabola opens upwards or downwards.

If the parabola opens upwards (when a > 0), the vertex represents the minimum value of the parabola. If a < 0, the parabola opens downwards, and the vertex represents the maximum value.

For a parabola in the standard form of a quadratic equation: y = ax² + bx + c

The coordinates of the vertex (x, y) are given by: x = −b/2a

Once you find the x-coordinate of the vertex, substitute it back into the equation to find the y-coordinate.

So, the vertex is located at: \left( \frac{-b}{2a}, y \right)

where y is the value of the function when x = \frac{-b}{2a}.

Formula

For the vertex form of the parabola, y = a(x - h)² + k, the coordinates (h, k) of the vertex are,

(h, k) = (-b/2a, -D/4a)

where,

a is the coefficient of x²,
b is the coefficient of x,
D = b² - 4ac is the discriminant of the standard form y = ax² + bx + c.

Derivation

Suppose we have a parabola with a standard equation as, y = ax² + bx + c.

This can be written as,

y - c = ax² + bx
y - c = a (x² + bx/a)

Adding and subtracting b²/4a² on the RHS, we get

y - c = a (x² + bx/a + b²/4a² - b²/4a²)
y - c = a ((x + b/2a)² - b²/4a²)
y - c = a (x + b/2a)² - b²/4a
y = a (x + b/2a)² - b²/4a + c
y = a (x + b/2a)² - (b²/4a - c)
y = a (x + b/2a)² - (b² - 4ac)/4a

We know, D = b² - 4ac, so the equation becomes,
y = a (x + b/2a)² - D/4a

Comparing the above equation with the vertex form y = a(x - h)² + k, we get
h = -b/2a and k = -D/4a

This derives the formula for coordinates of the vertex of a parabola.

Properties

• The vertex of every parabola is its turning point.
• The derivative of the parabola function at its vertex is always zero.
• A parabola that is either open at its top or bottom has a maxima or a minima at its vertex.
• The vertex of a left or right open parabola is neither a maxima nor a minima of the parabola.
• Vertex is the point of intersection between the parabola and its axis of symmetry.

Related Articles

• Standard Equation of a Parabola
• Focus and Directrix of a Parabola
• Applications of Parabola in Real-Life

Solved Examples of Vertex of a Parabola

Problem 1. Find the coordinates of the vertex for the parabola y = 2x² + 4x - 4.

Solution:

We have the equation as, y = 2x² + 4x - 4.

Here, a = 2, b = 4 and c = -4.

Now, it is known that the coordinates of the vertex are given by, (-b/2a, -D/4a) where D = b² - 4ac.
D = (4)² - 4 (2) (-4)
= 16 + 32
= 48

So, x - coordinate of vertex = -4/2(2) = -4/4 = -1.

y - coordinate of vertex = -48/4(2) = -48/8 = -6

Hence, the vertex of the parabola is (-1, -6).

Problem 2. Find the coordinates of the vertex for the parabola y = 3x² + 5x - 2.

Solution:

We have the equation as, y = 3x² + 5x - 2.

Here, a = 3, b = 5 and c = -2.

Now, it is known that the coordinates of the vertex are given by, (-b/2a, -D/4a) where D = b² - 4ac.
D = (5)² - 4 (3) (-2)
= 25 + 24
= 49

So, x - coordinate of vertex = -5/2(3) = -5/6
y - coordinate of vertex = -49/4(3) = -49/12

Hence, the vertex of the parabola is (-5/6, -49/12).

Problem 3. Find the coordinates of the vertex for the parabola y = 3x² - 6x + 1.

Solution:

We have the equation as, y = 3x² - 6x + 1.

Here, a = 3, b = -6 and c = 1.

Now, it is known that the coordinates of the vertex are given by, (-b/2a, -D/4a) where D = b² - 4ac.
D = (-6)² - 4 (3) (1)
= 36 - 12
= 24

So, x - coordinate of vertex = 6/2(3) = 6/6 = 1
y - coordinate of vertex = -24/4(3) = -24/12 = -2

Hence, the vertex of the parabola is (1, -2).

Problem 4. Find the coordinates of the vertex for the parabola y = 3x² + 8x - 8.

Solution:

We have the equation as, y = 3x² + 8x - 8.

Here, a = 3, b = 8 and c = -8.

Now, it is known that the coordinates of the vertex are given by, (-b/2a, -D/4a) where D = b² - 4ac.
D = (8)² - 4 (3) (-8)
= 64 + 96
= 160

So, x - coordinate of vertex = -8/2(3) = -8/6 = -4/3
y - coordinate of vertex = -160/4(3) = -160/12 = -40/3

Hence, the vertex of the parabola is (-4/3, -40/3).

Problem 5. Find the coordinates of the vertex for the parabola y = 6x² + 12x + 4.

Solution:

We have the equation as, y = 6x² + 12x + 4.

Here, a = 6, b = 12 and c = 4.

Now, it is known that the coordinates of the vertex are given by, (-b/2a, -D/4a) where D = b² - 4ac.
D = (12)² - 4 (6) (4)
= 144 - 96
= 48

So, x - coordinate of vertex = -12/2(6) = -12/12 = -1
y - coordinate of vertex = -48/4(6) = -48/24 = -2

Hence, the vertex of the parabola is (-1, -2).

Problem 6. Find the coordinates of the vertex for the parabola y = x² + 7x - 5.

Solution:

We have the equation as, y = x² + 7x - 5.

Here, a = 1, b = 7 and c = -5.

Now, it is known that the coordinates of the vertex are given by, (-b/2a, -D/4a) where D = b² - 4ac.
D = (7)² - 4 (1) (-5)
= 49 + 20
= 69

So, x - coordinate of vertex = -7/2(1) = -7/2
y - coordinate of vertex = -69/4(1) = -69/4

Hence, the vertex of the parabola is (-7/2, -69/4).

Problem 7. Find the coordinates of the vertex for the parabola y = 2x² + 10x - 3.

Solution:

We have the equation as, y = x2 + 7x - 5.

Here, a = 1, b = 7 and c = -5.

Now, it is known that the coordinates of the vertex are given by, (-b/2a, -D/4a) where D = b2 - 4ac.
D = (7)2 - 4 (1) (-5)
= 49 + 20
= 69

So, x - coordinate of vertex = -7/2(1) = -7/2
y - coordinate of vertex = -69/4(1) = -69/4

Hence, the vertex of the parabola is (-7/2, -69/4).

Practice Problems on Vertex of a Parabola

Question 1: Find the vertex of the parabola given by the equation y = 2x² - 4x + 1.

Question 2: Determine the vertex of the parabola y = -3x² + 6x - 2.

Question 3: Find the vertex of the quadratic function y = x² - 8x + 15.

Question 4: Determine the vertex of the parabola y = 5x² + 10x = 3.