Find the number of x-intercepts of the quadratic function, graph the parabola and verify it.
Problem 1 :
f(x) = x2 - x - 6
Problem 2 :
f(x) = x2 - 4
Problem 3 :
f(x) = (x - 2)2
Problem 4 :
f(x) = x2 - 4x + 7
1. Answer :
In the given quadratic function, put f(x) = 0.
x2 - x - 6 = 0
Comparing ax2 + bx + c = 0 and x2 - x - 6 = 0,
a = 1, b = -1 and c = -6
Value of b2 - 4ac :
= (-1)2 - 4(1)(-6)
= 1 + 24
= 25 > 0
Since b2 - 4ac > 0, solving the quadratic equation
x2 - x - 6 = 0
will yield has two real values for x.
Then there are two x-intercepts for the given quadratic function.
Graphing :
Solve x2 - x - 6 = 0 by factoring :
x2 - x - 6 = 0
(x + 2)(x - 3) = 0
x + 2 = 0 or x - 3 = 0
x = -2 or x = 3
Two x-intercepts are -2 and 3.
Comparing f(x) = ax2 + bx + c and f(x) = x2 - x - 6 = 0,
a = 1, b = -1 and c = -6
a = 1 > 0 ----> parabola opens up
x-coordinate of the vertex :
x = -b/2a
Substitute a = 1 and b = -1.
x = -(-1)/2(1)
x = 1/2
x = 0.5
y-coordinate of the vertex :
Substitute x = 0.5 in f(x) = x2 - x - 6.
f(0.5) = (0.5)2 - (0.5) - 6
= 0.25 - 0.5 - 6
= -6.25
Vertex of the parabola is (0.5, -6.25).
2. Answer :
In the given quadratic function, put f(x) = 0.
x2 - 4 = 0
Comparing ax2 + bx + c = 0 and x2 - 4 = 0,
a = 1, b = 0 and c = -4
Value of b2 - 4ac :
= 02 - 4(1)(-4)
= 0 + 16
= 16 > 0
Since b2 - 4ac > 0, solving the quadratic equation
x2 - 4 = 0
will yield two real values for x.
Then there are two x-intercepts for the given quadratic function.
Graphing :
Solve x2 - 4 = 0 by factoring :
x2 - 4 = 0
x2 - 22 = 0
Using the identity a2 - b2 = (a + b)(a - b),
(x + 2)(x - 2) = 0
x + 2 = 0 or x - 2 = 0
x = -2 or x = 2
Two x-intercepts are -2 and 2.
Comparing f(x) = ax2 + bx + c and f(x) = x2 - 4,
a = 1, b = 0 and c = -4
a = 1 > 0 ----> parabola opens up
x-coordinate of the vertex :
x = -b/2a
Substitute a = 1 and b = 0.
x = -(0)/2(1)
x = 0
y-coordinate of the vertex :
Substitute x = 0 in f(x) = x2 - 4.
f(0) = 02 - 4
= 0 - 4
= -4
Vertex of the parabola is (0, -4).
3. Answer :
f(x) = (x - 2)2
Write the given quadratic function in standard form :
f(x) = (x - 2)(x - 2)
f(x) = x2 - 2x - 2x + 4
f(x) = x2 - 4x + 4
In the given quadratic function above, put f(x) = 0.
x2 - 4x + 4 = 0
Comparing ax2 + bx + c = 0 and x2 - 4x + 4 = 0,
a = 1, b = -4 and c = 4
Value of b2 - 4ac :
= (-4)2 - 4(1)(4)
= 16 - 16
= 0
Since b2 - 4ac = 0, solving the quadratic equation
x2 - 4x + 4 = 0
will yield one real value for x.
Then there is one x-intercept for the given quadratic function.
Graphing :
Solve x2 - 4x + 4 = 0 by factoring :
x2 - 4x + 4 = 0
(x - 2)2 = 0
Taking square root on both sides,
x - 2 = 0
x = 2
x-intercept 2.
Comparing f(x) = ax2 + bx + c and f(x) = x2 - 4x + 4,
a = 1, b = -4 and c = 4
a = 1 > 0 ----> parabola opens up
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 f(x) = x2 - 4x + 4.
f(2) = 22 - 4(2) + 4
= 4 - 8 + 4
= 0
Vertex of the parabola is (2, 0).
4. Answer :
f(x) = x2 - 4x + 7
In the given quadratic function above, put f(x) = 0.
x2 - 4x + 7 = 0
Comparing ax2 + bx + c = 0 and x2 - 4x + 7 = 0,
a = 1, b = -4 and c = 7
Value of b2 - 4ac :
= (-4)2 - 4(1)(7)
= 16 - 28
= -12 < 0
Since b2 - 4ac < 0, solving the quadratic equation
x2 - 4x + 7 = 0
will yield no real value for x.
Then there is no x-intercept for the given quadratic function.
Graphing :
Solve x2 - 4x + 7 = 0 by factoring :
Already we know that b2 - 4ac < 0.
So, solving x2 - 4x + 7 = 0 will yield no real value for x.
There is no x-intercept. That is, the parabola will never intersect x-axis.
Comparing f(x) = ax2 + bx + c and f(x) = x2 - 4x + 7,
a = 1, b = -4 and c = 7
a = 1 > 0 ----> parabola opens up
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 f(x) = x2 - 4x + 7.
f(2) = 22 - 4(2) + 7
= 4 - 8 + 7
= 3
Vertex of the parabola is (2, 3).
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