Question

Prove that a>0 will always open the parabola upwards!

Answer 1

We have the equation as : \(\cfrac{Y}{a} = X^2\) , where : \(Y = y + \cfrac{D}{4a} \) and \(X = x + \cfrac{b}{2a}\)

Answer 2

i got the above equation : \(\cfrac{Y}{a} = X^2\) by shifting the origin to the point : \(\cfrac{-b}{2a} \space , \space \cfrac{-D}{4a}\) . It says that if a>0 the parabola will open upwards, can any one help me in proving this point?

Answer 3

From what you did, shifting the origin, if a is positive, then Y must always be positive since X^2 is positive for all X.

Answer 4

ok but how will it help me in prooving this?

Answer 5

Is there any use of calculus?

Answer 6

Hmm, I guess you could find the second derivative, and show that it's always positive, so the graph is concave up for all x.

Answer 7

yes! Can you show me a little of the work? Please?

I tried to differentiate : \(ax^2 + bx + c =0\) getting \(2ax + b = 0 \) or \(x= \cfrac{-b}{2a}\)

Answer 8

If you differentiate \[\large y= ax^2 + bx + c\] you'll get \[\large y'= 2ax + b\] differentiate again for the second derivative \[\large y''= 2a\] which is always positive for a>0, so y is concave up (opens upward)

Answer 9

Ok! Got it agent. Thanks a lot.

Answer 10

@ParthKohli do you mean if the acceleration is constant and positive? This type of equation has constant acceleration, not increasing.

Answer 11

you can prove this, by showing that function of type \(f(x)=ax^2+bx+c\) with \(a>0\) has a mínimum and no máximum. To do that find \(f'(x)\) which will be the candidate for the extremum. Later check the sign of \(f''(x)\) in the neighbourhood of this point to notice that it will be positive since \(f''(x)=a\), so the point in question will be a minimum

Answer 12

should be \(f''(x)=2a\), sry.

Answer 13

Thank you to @myko and @agent0smith , thanks a lot.