The standard form of a quadratic equation will be
y = ax^{2} + bx + c
Using the following two different ways, we can find the vertex of the parabola.
(i) Using completing the square
(ii) Using formula
Here we see, how to find vertex of the parabola from the standard form of the equation.
Name a, b and c for each parabola. Then find the vertex. Show all work.
Problem 1 :
f(x) = x² - 8x + 17
Solution :
Comparing f(x) = ax² + bx + c and f(x) = x² - 8x + 17
a = 1, b = -8 and c = 17
x-coordinate of the vertex :
x = -b/2a
Substitute a = 1 and b = -8.
x = -(-8)/2(1)
x = 8/2
x = 4
y-coordinate of the vertex :
Substitute x = 4 in f(x) = x² - 8x + 17
f(4) = 4² - 8(4) + 17
= 16 - 32 + 17
= 1
Vertex of the parabola is (4, 1).
Problem 2 :
f(x) = -x² - 2x - 2
Solution :
Comparing f(x) = ax² + bx + c and f(x) = -x² - 2x - 2
a = -1, b = -2 and c = -2
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 f(x) = -x² - 2x - 2
f(-1) = -(-1)² - 2(-1) - 2
= -1 + 2 - 2
= 1
Vertex of the parabola is (-1, 1).
Problem 3 :
f(x) = -x² + 6x - 8
Solution :
Comparing f(x) = ax² + bx + c and f(x) = -x² + 6x - 8
a = -1, b = 6 and c = -8
x - coordinate of the vertex :
x = -b/2a
Substitute a = -1 and b = 6.
x = -6/2(-1)
x = -6/-2
x = 3
y - coordinate of the vertex :
Substitute x
= 3 in f(x) = -x² + 6x - 8
f(3) = -3² + 6(3) - 8
= -9 + 18 - 8
= 1
Vertex of the parabola is (3, 1).
Problem 4 :
f(x) = -3x² + 6x
Solution :
Comparing f(x) = ax² + bx + c and f(x) = -3x² + 6x
a = -3, b = 6 and c = 0
x-coordinate of the vertex :
x = -b/2a
Substitute a = -3 and b = 6.
x = -6/2(-3)
x = -6/-6
x = 1
y-coordinate of the vertex :
Substitute x = 1 in f(x) = -3x² + 6x
f(1) = -3(1)² + 6(1)
= -3 + 6
= 3
Vertex of the parabola is (1, 3).
Problem 5 :
f(x) = -2x² - 16x - 31
Solution :
Comparing f(x) = ax² + bx + c and f(x) = -2x² - 16x - 31
a = -2, b = -16 and c = -31
x-coordinate of the vertex :
x = -b/2a
Substitute a = -2 and b = -16.
x = -(-16)/2(-2)
x = 16/-4
x = -4
y-coordinate of the vertex :
Substitute x = -4 in f(x) = -2x² - 16x - 31
f(-4) = -2(-4)² - 16(-4) - 31
= -32 + 64 - 31
= 64 - 63
= 1
Vertex of the parabola is (-4, 1).
Problem 6 :
f(x) = -1/2x² - 4x - 6
Solution :
Comparing f(x) = ax² + bx + c and f(x) = -1/2x² - 4x - 6
a = -1/2, b = -4 and c = -6
x-coordinate of the vertex :
x = -b/2a
Substitute a = -1/2 and b = -4.
x = -(-4)/2(-1/2)
x = 4/-1
x = -4
y-coordinate of the vertex :
Substitute x = -4 in f(x) = -1/2x² - 4x - 6
f(-4) = -1/2(-4)² - 4(-4) - 6
= -8 + 16 - 6
= 16 - 14
= 2
Vertex of the parabola is (-4, 2).
Problem 7 :
f(x) = -2x² + 12x - 22
Solution :
Comparing f(x) = ax² + bx + c and f(x) = -2x² + 12x - 22
a = -2, b = 12 and c = -22
x-coordinate of the vertex :
x = -b/2a
Substitute a = -2 and b = 12.
x = -12/2(-2)
x = -12/-4
x = 3
y-coordinate of the vertex :
Substitute x = 3 in f(x) = -2x² + 12x - 22
f(3) = -2(3)² + 12(3) - 22
= -18 + 36 - 22
= -40 + 36
= -4
Vertex of the parabola is (3, -4).
Problem 8 :
f(x) = x² - 8x + 20
Solution :
Comparing f(x) = ax² + bx + c and f(x) = x² - 8x + 20
a = 1, b = -8 and c = 20
x-coordinate of the vertex:
x = -b/2a
Substitute a = 1 and b = -8.
x = -(-8)/2(1)
x = 8/2
x = 4
y-coordinate of the vertex:
Substitute x = 4 in f(x) = x² - 8x + 20
f(4) = 4² - 8(4) + 20
= 16 - 32 + 20
= 4
Vertex of the parabola is (4, 4).
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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