How to find whether the parabola intersects x axis without graph :
Here we are going to how to find whether the parabola intersects x axis without graphing.
We may find number of points where the parabola intersects the x-axis, using the formula for discriminant.
Condition |
Number of intersecting points |
b^{2} − 4ac > 0 b^{2} − 4ac = 0 b^{2} − 4ac < 0 |
Intersects x-axis in two places Touches x-axis at only one point does not intersect x-axis |
Let us look into some examples to understand the above concept.
Question 1 :
Without sketching the graphs, find whether the graphs of the following functions will intersect the x-axis and if so in how many points.
(i) y = x^{2} + x + 2
Solution :
Discriminant = b^{2} − 4ac
a = 1, b = 1 and c = 2
b^{2} − 4ac = (1)^{2} − 4(1)(2)
= 1 - 8
= -7 < 0
Does not intersect the x-axis.
(ii) y = x^{2} − 3x − 7
Solution :
Discriminant = b^{2} − 4ac
a = 1, b = -3 and c = -7
b^{2} − 4ac = (-3)^{2} − 4(1)(-7)
= 9 + 28
= 36 > 0
Hence the curve intersects x-axis at two points.
(iii) y = x^{2} + 6x + 9
Solution :
Discriminant = b^{2} − 4ac
a = 1, b = 6 and c = 9
b^{2} − 4ac = (6)^{2} − 4(1)(9)
= 36 - 36
= 0
Hence the curve does not intersect x-axis at any point.
Question 2 :
Write f(x) = x^{2} + 5x + 4 in completed square form.
Solution :
= x^{2} + 5x + 4
Multiply and divide the coefficient of x by 2.
= x^{2} + (2/2) ⋅ 5 ⋅ x + 4
= x^{2} + 2 ⋅ x ⋅ (5/2) + (5/2)^{2} - (5/2)^{2 }+ 4
= (x + (5/2))^{2} - (25/4)^{ }+ 4
= (x + (5/2))^{2} + (16 - 25)/4
= (x + (5/2))^{2} + (-9)/4
= (x + (5/2))^{2} - (3/2)^{2}
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