clara1223
  • clara1223
How to graph f(x) = ax^2 + bx + c when a<0, b^2-4ac=0?
Mathematics
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clara1223
  • clara1223
How to graph f(x) = ax^2 + bx + c when a<0, b^2-4ac=0?
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
it looks something like this|dw:1438599255017:dw|
clara1223
  • clara1223
@saseal Could you please explain?
anonymous
  • anonymous
this is a quadratic equation, which makes it a parabola. since a for ax^2 is greater than 0 the parabola opens upward. if b^2-4ac=0, the curve theres only 1 x-intercept, instead of 2 roots

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anonymous
  • anonymous
c is always your y-intercept
anonymous
  • anonymous
you can understand that?
clara1223
  • clara1223
@saseal from b^2-4ac how did you find that there is only one root and where c is?
anonymous
  • anonymous
|dw:1438600059555:dw||dw:1438600096383:dw||dw:1438600146075:dw|
anonymous
  • anonymous
if you remember the y=mx+c equation, c is the y-intercept. thats how we know. c is always the constant number
clara1223
  • clara1223
how did you find that the equation moves to the left?
anonymous
  • anonymous
this is something you need to figure it out by plugging the numbers in after you solve the equation
clara1223
  • clara1223
There are no numbers provided in the question.
anonymous
  • anonymous
then x could be anywhere on the line, but your curve should only touch the x-axis once
anonymous
  • anonymous
the curve can be anywhere along the x-axis |dw:1438600611313:dw|

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