anonymous
  • anonymous
how to define this equation by switching operations to g-^1 g(x)=-2(x-1)^2 + 7
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
oh yeah but how do we find the x intercepts of that inverse function?
anonymous
  • anonymous
and how come there's a negative sign here => -(x-7)

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anonymous
  • anonymous
the x-intercept is the point such that it hits the x axis which is the line where y=0 so to find the x-intercept of g^-1 we set y=g^-1 and solve y = 1 +- sqrt((7 - x)/2) 0 = 1+- sqrt((7 - x)/2) -1 = +- sqrt((7 - x)/2) 1 = (7 - x)/2 2 = 7 - x -5 = -x x = 5 so the x-intercept is (5, 0) the reason there is a negative in that step before the (x - 7) is because i divided by -2 and transferred that negative to the numerator because it's standard for the denominator to be positive
anonymous
  • anonymous
thank you very much
anonymous
  • anonymous
and how do we find the vertex of it?
anonymous
  • anonymous
of g^-1? well, i'd put it into horizontal parabola vertex form: x = a(y - h)^2 + k so x = -2(y - 1)^2 + 7 this makes it easy to find the vertex, which is (h,k). if you look at the equation, we see h = 1 and k = 7 so the vertex of g^-1 is (1,7)
anonymous
  • anonymous
http://hotmath.com/hotmath_help/topics/vertex-of-a-parabola.html
anonymous
  • anonymous
the vertex is in the exact spot as the original equation but not the inverse function >_>
anonymous
  • anonymous
yeah sorry my bad the vertex is actually (k,h) for horizontal. if you rewrite the vertex form as: x - k = a(y - h)^2 you just look at which is subtracted from what variable to determine what coordinate it designates for the vertex. since it's (k,h) you should have (7,1). my bad
anonymous
  • anonymous
the vertex coordinates swap between g and g^-1 which makes sense because you're swapping the x,y values
anonymous
  • anonymous
ohh i see
anonymous
  • anonymous
but for the x intercept im supposed to also have 2 coordinates is 5,0 one of them?

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