Case Study Based:
 Parabola : A parabola is the graph that
results from plx) = ax2+ bx2+c
Parabola is symmetric about a vertical
line which is called axls of symmetry
The axis of symmetry runs through the maximum and minimum point of the
parabola which is called the vertex.


(a) lf the suspension bridge is represented by x2-2x-3, then it's zeroes are:
(i)2,-1       (ii)3,-1
(iii)-1,-3    (iv)-1,4

(b)lf the suspension bridge is represented graphically, .then the number of zeroes of quadratic equation of parabolic cable is equal to the number of points where the graph of cable: 
(i) intersect y-axis                 (ii) intersects x-axis
(iii)either x-axis or y-axis     (iv)none of these

(c) The cable of suspension bridge is represented graphically as shown in the given figure.

Find the number of its zeroes
(i)1  (ii)0
(iii)2 (iv)3

(d) Graph of a quadratic polynomial is a
(i)straight line  (ii)circle
(iii)parabola      (iv)ellipse
e) The representation of cables of suspension bridge whose one zero is 5 and sum of the zeroes is 0, is
(i)x2  - 5x + 25 (ii) x- 25
(iii)x- 5          (iv) x- √5

 

Dear Student,
The solution/s to your  1st and 5th query have been provided below:
a.Finding Zeroes
x2 - 2x -3 = 0
Solving by splitting the middle term method
x2 - 3x + x - 3 = 0
x(x - 3) + (x - 3) = 0
(x-3) (x + 1) = 0
Therefore,
x=3,-1
So, (ii) is correct

e.Sum of Zeros = 0
First Zero + Second Zero = 0
5 + Second Zero = 0
Second Zero = -5
The Required polynomial is
p(x) = x2 - (Sum of Zeroes)x + Product of Zeroes
= x2 - 0x + (5 x-5)
= x2 - 25
So, (ii) is correct

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