The vertex of a parabola is the point where the parabola crosses its axis of symmetry.
The vertex of a parabola is the highest or lowest point which is also known as maximum or minimum value.
If a parabola opens up, the vertex will be the lowest point and if it opens down, the vertex will be the highest point of the parabola.
Equation of a parabola in standard form :
y = ax2 + bx + c
Equation of a parabola in vertex form :
y = a(x - h)2 + k
where (h, k) is the vertex.
In both standard form and vertex form, if
a < 0 ----> parabola opens down
a > 0 ----> parabola opens up
When the equation of a parabola is given in standard form, you can find the vertex of the parabola in two methods.
Method 1 :
Let the equation of a parabola be y = ax2 + bx + c.
You can use the formula given below to find the x-coordinate of the vertex.
x = ⁻ᵇ⁄₂ₐ
After having found the x-coordinate, you can substitute it into the equation of the parabola and find the y-coordinate of the vertex.
Method 2 :
You can write the given equation of a parabola in vertex form and find the vertex.
In each case, find the vertex of the parabola :
Example 1 :
y = x2 - 2x - 5
Solution :
Method 1 :
Comparing y = ax2 + bx + c and y = x2 - 2x - 5,
a = 1, b = -2 and c = -5
x-coordinate of the vertex :
x = ⁻ᵇ⁄₂ₐ
Substitute a = 1 and b = -2.
x = ⁻⁽⁻²⁾⁄₂₍₁₎
x = ²⁄₂
x = 1
y-coordinate of the vertex :
Substitute x = 1 in y = x2 - 2x - 5.
y = 12 - 2(1) - 5
y = 1 - 2 - 5
y = -6
Vertex of the parabola is (1, -6)
Method 2 :
Write the given equation of parabola in vertex form.
y = x2 - 2x - 5
y = x2 - 2(x)(1) + 12 - 12 - 5
Using the identity (a - b)2 = a2 - 2ab + b2,
y = (x - 1)2 - 12 - 5
y = (x - 1)2 - 1 - 5
y = (x - 1)2 - 6
Comparing y = a(x - h)2 + k and y = (x - 1)2 - 6,
h = 1 and k = -6
Vertex of the parabola :
(h, k) = (1, -6)
Example 2 :
y = -x2 - 14x - 59
Solution :
Method 1 :
Comparing y = ax2 + bx + c and y = -x2 - 14x - 59,
a = -1, b = -14 and c = -59
x-coordinate of the vertex :
x = ⁻ᵇ⁄₂ₐ
Substitute a = -1 and b = -14.
x = ⁻⁽⁻¹⁴⁾⁄₂₍₋₁₎
x = ¹⁴⁄₋₂
x = -7
y-coordinate of the vertex :
Substitute x = -7 in y = -x2 - 14x - 59.
y = -(-7)2 - 14(-7) - 59
y = -49 + 98 - 59
y = -10
Vertex of the parabola is (-7, -10)
Method 2 :
Write the given equation of parabola in vertex form.
y = -x2 - 14x - 59
y = -1(x2 + 14x) - 59
y = -1[x2 + 2(x)(7) + 72 - 72] - 59
Using the identity (a + b)2 = a2 + 2ab + b2,
y = -1[(x + 7)2 - 72] - 59
y = -1[(x + 7)2 - 49] - 59
y = -1(x + 7)2 + 49 - 59
y = -1(x + 7)2 - 10
The vertex form equation y = -1(x + 7)2 - 10 can be written as
y = -1[x - (-7)]2 + (-10)
Comparing the above equation and y = a(x - h)2 + k,
h = -7 and k = -10
Vertex of the parabola :
(h, k) = (-7, -10)
Example 3 :
y = x2 + 4x
Solution :
Method 1 :
Comparing y = ax2 + bx + c and y = x2 + 4x,
a = 1, b = 4 and c = 0
x-coordinate of the vertex :
x = ⁻ᵇ⁄₂ₐ
Substitute a = 1 and b = 4.
x = ⁻⁴⁄₂₍₁₎
x = ⁻⁴⁄₂
x = -2
y-coordinate of the vertex :
Substitute x = -2 in y = x2 + 4x.
y = (-2)2 + 4(-2)
y = 4 - 8
y = -4
Vertex of the parabola is (-2, -4)
Method 2 :
Write the given equation of parabola in vertex form.
y = x2 + 4x
y = x2 + 2(x)(2) + 22 - 22
Using the identity (a + b)2 = a2 + 2ab + b2,
y = (x + 2)2 - 22
y = (x + 2)2 - 4
The vertex form equation y = (x + 2)2 - 4 can be written as
y = [x - (-2)]2 + (-4)
Comparing the above equation and y = a(x - h)2 + k,
h = -2 and k = -4
Vertex of the parabola :
(h, k) = (-2, -4)
Equation of a parabola in standard form :
x = ay2 + by + c
Equation of a parabola in vertex form :
x = a(y - k)2 + h
where (h, k) is the vertex.
In both standard form and vertex form, if
a < 0 ----> parabola opens to the left
a > 0 ----> parabola opens up to the right
When the equation of a parabola is given in standard form, you can find the vertex of the parabola in two methods.
Method 1 :
Let the equation of a parabola be x = ay2 + by + c.
You can use the formula given below to find the y-coordinate of the vertex.
y = ⁻ᵇ⁄₂ₐ
After having found the y-coordinate, you can substitute it into the equation of the parabola and find the x-coordinate of the vertex.
Method 2 :
You can write the given equation of a parabola in vertex form and find the vertex.
In each case, find the vertex of the parabola :
Example 4 :
x = y2 - 2y - 5
Solution :
Method 1 :
Comparing x = ay2 + by + c and x = y2 - 2y - 5,
a = 1, b = -2 and c = -5
y-coordinate of the vertex :
y = ⁻ᵇ⁄₂ₐ
Substitute a = 1 and b = -2.
y = ⁻⁽⁻²⁾⁄₂₍₁₎
y = ²⁄₂
y = 1
x-coordinate of the vertex :
Substitute y = 1 in x = y2 - 2y - 5.
x = 12 - 2(1) - 5
x = 1 - 2 - 5
x = -6
Vertex of the parabola is (-6, 1)
Method 2 :
Write the given equation of parabola in vertex form.
x = y2 - 2y - 5
x = y2 - 2(y)(1) + 12 - 12 - 5
Using the identity (a - b)2 = a2 - 2ab + b2,
x = (y - 1)2 - 12 - 5
x = (y - 1)2 - 1 - 5
x = (y - 1)2 - 6
Comparing x = a(y - k)2 + h and y = (x - 1)2 - 6,
h = -6 and k = 1
Vertex of the parabola :
(h, k) = (-6, 1)
Example 5 :
x = -y2 - 14y - 59
Solution :
Method 1 :
Comparing x = ay2 + by + c and x = -y2 - 14y - 59,
a = -1, b = -14 and c = -59
x-coordinate of the vertex :
y = ⁻ᵇ⁄₂ₐ
Substitute a = -1 and b = -14.
y = ⁻⁽⁻¹⁴⁾⁄₂₍₋₁₎
y = ¹⁴⁄₋₂
y = -7
y-coordinate of the vertex :
Substitute y = -7 in x = -y2 - 14y - 59.
x = -(-7)2 - 14(-7) - 59
x = -49 + 98 - 59
x = -10
Vertex of the parabola is (-10, -7)
Method 2 :
Write the given equation of parabola in vertex form.
x = -y2 - 14y - 59
x = -1(y2 + 14y) - 59
x = -1[y2 + 2(y)(7) + 72 - 72] - 59
Using the identity (a + b)2 = a2 + 2ab + b2,
x = -1[(y + 7)2 - 72] - 59
x = -1[(y + 7)2 - 49] - 59
x = -1(y + 7)2 + 49 - 59
x = -1(y + 7)2 - 10
The vertex form equation x = -1(y + 7)2 - 10 can be written as
x = -1[y - (-7)]2 + (-10)
Comparing the above equation and x = a(y - k)2 + h,
h = -10 and k = -7
Vertex of the parabola :
(h, k) = (-10, -7)
Example 6 :
x = y2 + 4y
Solution :
Method 1 :
Comparing x = ay2 + by + c and x = y2 + 4y,
a = 1, b = 4 and c = 0
x-coordinate of the vertex :
y = ⁻ᵇ⁄₂ₐ
Substitute a = 1 and b = 4.
y = ⁻⁴⁄₂₍₁₎
y = ⁻⁴⁄₂
y = -2
y-coordinate of the vertex :
Substitute y = -2 in x = y2 + 4y.
x = (-2)2 + 4(-2)
x = 4 - 8
x = -4
Vertex of the parabola is (-4, -2).
Method 2 :
Write the given equation of parabola in vertex form.
x = y2 + 4y
x = y2 + 2(y)(2) + 22 - 22
Using the identity (a + b)2 = a2 + 2ab + b2,
x = (y + 2)2 - 22
x = (y + 2)2 - 4
The vertex form equation x = (y + 2)2 - 4 can be written as
x = [y - (-2)]2 + (-4)
Comparing the above equation and x = a(y - k)2 + h,
h = -4 and k = -2
Vertex of the parabola :
(h, k) = (-4, -2)
Intercept form equation of a parabola :
y = a(x - p)(x - q)
Substitute y = 0 to find the x-intercepts of the parabola above,
a(x - p)(x - q) = 0
Divide both sides by a.
(x - p)(x - q) = 0
x - p = 0 or x - q = 0
x = p or x = q
The two x-intercepts of the parabola y = a(x - p)(x - q) are p and q.
To find the x-coordinate of the vertex, find the average of the two x-intercepts.
After having found the x-coordinate, you can substitute it into the equation of the parabola and find the y-coordinate of the vertex.
Example 7 :
y = -3(x + 2)(x - 6)
Solution :
y = -3(x + 2)(x - 6)
Substitute y = 0 to find the x-intercepts of the parabola above.
-3(x + 2)(x - 6) = 0
Divide both sides by -3.
-3(x + 2)(x - 6) = 0
x + 2 = 0 or x - 6 = 0
x = -2 or x = 6
x-coordinate of the vertex :
x = ⁽⁻² ⁺ ⁶⁾⁄₂
x = ⁴⁄₂
x = 2
y-coordinate of the vertex :
Substitute x = 2 in y = -3(x + 2)(x - 6).
y = -3(2 + 2)(2 - 6)
y = -3(4)(-4)
y = 48
Vertex of the parabola is (2, 48).
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