anonymous
  • anonymous
how do you solve a parabola? y=x^2-4x+3 i know you find the intercepts, y=3 and x=3, 1. but what next?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
what do you want done? what do you mean to "solve a parabola"?
anonymous
  • anonymous
like, graph a parabola... i think
anonymous
  • anonymous
you want to find the axis of symmetry and the vertex

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anonymous
  • anonymous
how do i do that?
anonymous
  • anonymous
in the form \[ax^2+bx+c\] the axis of symmetry is \[x=\frac{-b}{2a}\] the vertex is \[(\frac{-b}{2a}, f(\frac{-b}{2a}))\]
anonymous
  • anonymous
so y=x^2-4x+3 would become x=4/2? for the axis of symmetry
anonymous
  • anonymous
and for the vertex it would be 4/2 times f(4/2)? what does the f mean?
anonymous
  • anonymous
as in f(x) plug in -b/(2a) into the equation of the parabola to get the y-coordinate
anonymous
  • anonymous
so vertex is at (2, -1)
anonymous
  • anonymous
it is a point
anonymous
  • anonymous
put 4/2 (-b/2a) where f(x) is?
anonymous
  • anonymous
no.. f(x)=x^2-4x+3 this is the equation of your parabola the axis of symmetry is the vertical line that passes through the x-coordinate of your vertex To get the y-coordinate, plug in the x-coordinate into f(x) so f(2)=-1
anonymous
  • anonymous
where'd you get f(2)=-1?
anonymous
  • anonymous
plug 2 into f(x) f(2)=2^2-4(2)+3=4-8+3=-1
anonymous
  • anonymous
oh ok alright so the f(4/2) = 2^2-4(2)+3 so f(2) = 4-8 +3 so f(2) = -1. alright, so that's the y coordinate?
anonymous
  • anonymous
yes, the y-coordinate of the vertex the vertex is the point (2,-1)
anonymous
  • anonymous
ok. so let me just review here. the x coordinate to the vertex is -b/2a, which we then plug in for f(x) and solve for the y coordinate. where do the two side points come from?
anonymous
  • anonymous
what 2 side points?
anonymous
  • anonymous
the vertex is the middle one, but there have to be two other points, don't there? it's not a parabola with one point
anonymous
  • anonymous
there are an infinite number of points on a parabola, not just 3
anonymous
  • anonymous
you want the x-intercepts, the y-intercept, the vertex, and the axis of symmetry. a few other points won't hurt either
anonymous
  • anonymous
yeah, but you are only require to graph 3... so where do i get the other 2? :o
anonymous
  • anonymous
you already found the x-intercepts, (3, 0) and (1, 0)
anonymous
  • anonymous
the y-intercept is (0, 3) not y=3
anonymous
  • anonymous
ohhhh ok. so the three points are (3,0), (1,0), and (2, -1)
anonymous
  • anonymous
you could use those, i'd also use a 4th point for the y-intercept of (0,3)
anonymous
  • anonymous
oh ok. thank you sooooo much. you're a life saver!
anonymous
  • anonymous
yw

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