FIND THE VERTEX FORM OF THE QUADRATIC FUNCTION
We may have quadratic function in two different forms,
i) leading coefficient of x2 is 1.
ii) leading coefficient of x2 is not 1.
Writing vertex form of quadratic function which has the leading coefficient 1 :
- Write the coefficient of x as multiple of 2.
- By representing the given in the form
a2 + 2ab + b2 (or) a2 - 2ab + b2
we can get vertex form which is y = a(x - h)2 + k
Where a = 1
Writing vertex form of quadratic function which has the leading coefficient that is not equal to 1 :
- Factor the leading coefficient from x2 term and x term.
- Write the coefficient of x as multiple of 2.
- By representing the given in the form
a2 + 2ab + b2 (or) a2 - 2ab + b2
we can get vertex form which is y = a(x - h)2 + k
Where a is not equal to 1.
Use the information provided to write the vertex form equation of each parabola.
Problem 1 :
y = x2 - 4x + 5
Solution:
Using completing the square method,
y = x2 - 2x(2) + 22 - 22 + 5
y = (x - 2)2 - 22 + 5
y = (x - 2)2 + 1
By comparing this with the vertex form of parabola, we get
(h, k) = (2, 1)
Problem 2 :
y = x2 - 16x + 70
Solution:
Using completing the square method,
y = x2 - 2x(8) + 82 - 82 + 70
y = (x - 8)2 - 82 + 70
y = (x - 8)2 + 6
By comparing this with the vertex form of parabola, we get
(h, k) = (8, 6)
Problem 3 :
y = x2 - 4x + 2
Solution:
Using completing the square method,
y = x2 - 2x(2) + 22 - 22 + 2
y = (x - 2)2 - 22 + 2
y = (x - 2)2 - 2
By comparing this with the vertex form of parabola, we get
(h, k) = (2, -2)
Problem 4 :
y = -3x2 + 48x - 187
Solution:
Using completing the square method,
y = -3x2 + 48x - 187
y = -3(x2 - 16x) - 187
y = -3(x2 - 2x(8) + 82 - 82) - 187
y = -3((x - 8)2 - 64) - 187
= -3(x - 8)2 + 192 - 187
y = -3(x - 8)2 + 5
By comparing this with the vertex form of parabola, we get
(h, k) = (8, 5)
Problem 5 :
y = -2x2 - 12x - 12
Solution:
Using completing the square method,
y = -2x2 - 12x - 12
y = -2(x2 + 6x) - 12
= -2(x2 + 2(x)(3) + 32 - 32) - 12
= -2((x + 3)2 - 9) - 12
= -2(x + 3)2 + 18 - 12
y = -2(x + 3)2 + 6
By comparing this with the vertex form of parabola, we get
(h, k) = (-3, 6)
Problem 6 :
y = 3x2 + 18x + 18
Solution:
y = 3x2 + 18x + 18
= 3(x2 + 6x) + 18
= 3(x2 + 2(x)(3) + 32 - 32) + 18
= 3((x + 3)2 - 9) + 18
= 3(x + 3)2 - 27 + 18
y = 3(x + 3)2 - 9
By comparing this with the vertex form of parabola, we get
(h, k) = (-3, -9)
Problem 7 :
y = 2x2 + 3
Solution:
y = 2x2 + 3
y = 2(x - 0)2 + 3
By comparing this with the vertex form of parabola, we get
(h, k) = (0, 3)
Problem 8 :
y = 4x2 - 56x + 200
Solution:
y = 4x2 - 56x + 200
y = 4(x2 - 14x) + 200
= 4(x2 - 2(x)(7) + 72 - 72) + 200
= 4((x - 7)2 - 49) + 200
= 4(x - 7)2 - 196 + 200
y = 4(x - 7)2 + 4
By comparing this with the vertex form of parabola, we get
(h, k) = (7, 4)
Problem 9 :
y = -8x2 - 80x - 199
Solution:
y = -8x2 - 80x - 199
y = -8(x2 + 10x) - 199
= -8(x2 + 2(x)(5) + 52 - 52) - 199
= -8((x + 5)2 - 25) - 199
= -8(x + 5)2 + 200 - 199
y = -8(x + 5)2 + 1
By comparing this with the vertex form of parabola, we get
(h, k) = (-5, 1)
Problem 10 :
y = -2x2 + 20x - 52
Solution:
y = -2x2 + 20x - 52
y = -2(x2 - 10x) - 52
= -2(x2 - 2(x)(5) + 52 - 52) - 52
= -2((x - 5)2 - 25) - 52
= -2(x - 5)2 + 50 - 52
y = -2(x - 5)2 - 2
By comparing this with the vertex form of parabola, we get
(h, k) = (5, -2)
Recent Articles
-
Finding Range of Values Inequality Problems
May 21, 24 08:51 PM
Finding Range of Values Inequality Problems -
Solving Two Step Inequality Word Problems
May 21, 24 08:51 AM
Solving Two Step Inequality Word Problems -
Exponential Function Context and Data Modeling
May 20, 24 10:45 PM
Exponential Function Context and Data Modeling