Functions doubt
In the line I have highlighted, why did we assume the leading coefficient of the quadratic to be greater than zero?
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@Apu
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the graph wont cut the x axis :)
Ye it won't cut if the leading coefficient (x^2 ka coefficient) is greater than zero
Why is it greater than zero in this case?
That's the part I'm confused on
I understand the result applied although don't applied how the conditions of the result are met
Because thats the way we get an upward parabola and you need the quadratic to be greater than pr equal to zero hence a>0 is the only option
What if the coefficient was negative ? Can D somehow be less than zero in that case?
y=ax^2+bx+c is an equation of parabola on y axis with two possible orientations. Upward opening and downward opening. A layman explanation without hardcore proof is that Since x^2 will grow faster than the linear part, the coefficient of x^2 determines the sign at a point more or less. So for upward opening and positive values a should be greater
D depends on values of a,b,c. D can be less than zero in case a is negative too
But then the parabola will be downward facing without ever intersecting the x axis
Which means the quadratic will always have a negative value
But we dont want that right? We want greater than 0 values
D describes whether the quadratic will have any roots or not. Coefficient determines how the parabola will behave at infinity
Together D and a defines whether the quadratic will be always positive or always negative
Oh alright
Thanks
You sure you understood the point?
Yup :D
Sawal karte karte brain fuse ho gya tha
Aapne basics yaad dilaye toh samaj aa gya
Cool. Mark it as solved as mentioned here
+solved @Satya S
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