If the highway overpass is represented by `x^2-2x-8`.Then its zeroes are (i) (2,-4) (ii) (4,-2) (iii) (-2,-2) (iv) (-4,-4)

(i) x^(4) (ii) x^(-3) (iii) 3x^(2)

Case Study Based- 3 Applications of Parabolas-Highway Overpasses/Underpasses A highway underpass is parabolic in shape. Parabola A parabola is the graph that results from p(x) = ax^(2)+bx+c Parabolas are symmetric about a vertical line known as the Axis of Symmetry. The Axis of Symmetry runs through the maximum or minimum point of the parabola which is called the Vertex. If the highway overpass is represented by x^(2) - 2x - 8 . Then its zeroes are

In which quadrant does the given points lie (i) (4,-2) (ii) (-2,-2) (iii) (4,3) (iv) (-3,4)

The vertex of the parabola x^(2)+8x+12y+4=0 is (i) (-4,1) (ii) (4,-1)( iii) (-4,-1)

Locate the following points (i) (1,-1,3) " " (ii) (-1,2,4) (iii) (-2,-4,-7) " " (iv) (-4,2,-5)

The representation of Highway Underpass whose one zero is 6 and sum of the zeroes is 0, is (i) x^(2) -6x+2 (ii) x^(2) -36 (iii) x^(2)-6 (iv) x^(2)-3

Factorize (i) (x^4+4) (ii) (x^2+(4)/(x^2)) (iii) (x^4+(1)/(x^4)+1)

Check which of the following are solutions of the equations x-2y=4 and which are not: (i) (0,2) (ii) (2,0) (iii) (4,0) (iv) (sqrt(2),4sqrt(2)) (v) (1," "1)

Factorize (i) (x^4+3x^2+4) (ii) (x^4+5x^2+9)