y = ax^2 + bx + c
those above answers are right as for a and c.. but i would rather try to prove:
First c is the easiest to target
in simple
y= ax^2+bx+c
x=0 and y=c
so, in c always determines the y intercept.
now going to try a and b:
awesome... i figured it out = P
y= ax^2 + bx + c
= a(x^2 + bx) + c
now convert this to that form: a(x-h)^2 + k
=a(x^2+bx+(b^2/4))-((ab^2)/4)+c
=a(x+0.5b)^2 + (c- a.(b/2)^2)
so, now when x=-0.5b
all the first part equals to 0.. and then
y would be = to (c- a.(b/2)^2)
and, then this can only be the highest value or the lowest
because of a is negative that means this would be the highest
since whatever this (c- a.(b/2)^2) is, it would be subtracted by that
a(x+0.5b)^2 part
and if a is positive that means, when that first part is not equal to 0.. all those values would add to the second part.. and then that means y would be maximum only when the first part equals to 0
so we can conclude that the second part determines the vertex, (the maximum and the minimum value of y).
so, vertex is when x= - 0.5b
so that means b affects the x value in the vertex, like changing b would shift the vertex.
x) If the value of b is increased, the vertex would move towards left
and, if the value of is reduced, the vertex would move to the right side ...
and now let's see the y value in the vertex
y=(c- a.(b/2)^2)
Increasing c value means, the vertex would go up, and that means graph would be shifted up, if c value is increased.
Increasing a and b values means the graph would go down in y terms...
***************************************************************
Now going to see how the graph shape is affected:
y=a(x+0.5b)^2 + (c- a.(b/2)^2)
we know the second part (c- a.(b/2)^2) is a constant.. and it has no affect on how graph widens if those values are changed.
only the first part affects the shape of the graph:
a(x+0.5b)^2
Increasing a value would mean the y value would raise more and more when x value is changed.. so the graph would be skinny..
and b value also do the same.. but not as much as "a" do..
if u know calculus.. then here's that proof:
y=a(x+0.5b)^2
dy/dx=2a(x+0.5b)
so change in y value with respect to the change in the x value is also dependent upon b..
hope this helps..
But i am not good at communicating ideas.. =q .. so wondering..
2006-12-31 11:31:22
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answer #1
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answered by Anonymous
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f(x) = ax^2 + bx + c = 0
If a > 0, then the parabola will be concave downward and will have a minimum value.
If a <0 then the parabola will be concave upward and will have a maximum value.
The axis of symmetry will be x = -b/(2a).
The focus and the vertex will lie on the axis of symmetry.
If a<0 and b>0, the axis of symmetry will be parallel to the y-axis and to the right of it.
If b^2 - 4ac < 0, then f(x) has no real roots. The roots will be imaginary and f(x) will lie either entirely above the x-axis or entirely below the x-axis.
c will move the graph up or down c units
2006-12-31 12:19:14
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answer #2
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answered by ironduke8159 7
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There is another way to look at this. At x = 0, y = c, so the curve crosses the y axis at y = c. The minimum value of the curve occurs when dy/dx = 0, or 2ax+b=0, or where x =-b/2a. For very large values of x, the equation approximates y = ax^2 which is a concave-upward parabola, which narrows as a increases.
2016-05-23 01:13:24
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answer #3
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answered by Anonymous
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It's better to consider the vertex form of quadratic function:
f(x) = a(x-h)^2 + k
When a increases, the graph becomes narrower.
When h increases, the graph moves to the right.
When k increases, the graph moves to the left.
Now, can you find the opposite trend?
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By the way, comparing to f(x) = ax^2+bx+c, we have
a = q, h = (-b/2a), k = c
Therefore, whether the graph moves to the right or to the left depends on the change of -b/(2a)
2006-12-31 10:26:29
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answer #4
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answered by sahsjing 7
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a changes, graph will get either wider or narrower
b changes, the vertex will be lower or higher
c changes, the y-int changes
2006-12-31 10:04:07
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answer #5
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answered by 7
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