Question

y=ax²+bx+c is a equation of ________(fill in the blanks)
(1) write the points where the above graph cuts y=0line.
(2) write the point where the equation cut y axis.
(3) write the equation of straight line(axis of symmetry) that passes through the vertex.
(4) write the point of vertex .
(5) draw a suggestive graph on a plain paper
(a) where a is negative
(b) where a and b have different sign​

Answer 1

Answer:

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Answer 2

Answer:

step by step explanation:

given : y=ax²+bx+c

1) to know the point where y=0 cuts on above equation, substitute y=0 in it.

ax²+bx+c =0

from the formula of roots of a quadratic equation,

X = (-b+√(b²-4ac))/2a or x= (-b-√(b²-4ac))/2a

so points are ((-b+√(b²-4ac))/2a, 0) and ((-b-√(b²-4ac))/2a, 0)

2) to know the point where the equation cut y axis, substitute X=0. because on y axis , X will always be 0.

y=a(0)²+b(0)+c

y= c

it will cut only at one point i.e. (0,c)

3) to get the equation of straight line(axis of symmetry) that passes through the vertex, let us find vertex point first.

formula of vertex = [(-b/2a), y(-b/2a)]

y(-b/2a) is nothing but substitute -b/2a in given eqn.

and line passes through it will be y = y(-b/2a)

y = a(-b/2a) + b(-b/2a) +c

= -b/2 -b²/2a +c

y = (2ac - ab -b²)/2a

4) point of vertex from above question is [(-b/2a), (2ac - ab -b²)/2a]

5) check the attachment, sorry for bad hand writing

hope it helps, if yes make it brainliest.