anonymous
  • anonymous
https://cdn.ple.platoweb.com/EdAssets/560702b9dc1348a5974f50f34a330372?ts=635361559291470000 The graph represents the function f(x) = x2 + 3x + 2. If g(x) is the reflection of f(x) across the x-axis, g(x) = . (Write the function in standard form. Use ^ to indicate an exponent.)
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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jim_thompson5910
  • jim_thompson5910
hint: if you reflect f(x) over the x axis to get g(x), then g(x) = -f(x)
anonymous
  • anonymous
im still confused haha
jim_thompson5910
  • jim_thompson5910
let's say f(x) = x^2 - 6x + 10 multiply both sides by -1 to get -1*f(x) = -1*(x^2 - 6x + 10) -1*f(x) = -1*x^2 -1*(-6x) - 1*10 ... distribute -f(x) = -x^2 + 6x - 10 ------------------------------------------ If you compare f(x) = x^2 - 6x + 10 to -f(x) = -x^2 + 6x - 10, you'll find that all of the signs have been flipped. Eg: +x^2 goes to -x^2 or -6x turns into +6x So if f(x) = x^2 - 6x + 10, and we flip it over the x axis to get g(x), then g(x) = -f(x) g(x) = -x^2 + 6x - 10 this isn't the answer as it's just an example

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anonymous
  • anonymous
so then would i do g(x)=x^2 + 3x + 2?
jim_thompson5910
  • jim_thompson5910
that's f(x) to find g(x), you simply swap all of the signs on f(x)
anonymous
  • anonymous
so then it would be g(x)=x^2-3x-2? right? sorry im pretty bad at algebra
jim_thompson5910
  • jim_thompson5910
don't forget about the first term x^2
anonymous
  • anonymous
so that needs to have a negative sign in front of it also?
jim_thompson5910
  • jim_thompson5910
yes because in f(x), the x^2 is like +x^2
anonymous
  • anonymous
oh okay haha so then what do i do next?
jim_thompson5910
  • jim_thompson5910
the +x^2 just turns into -x^2
jim_thompson5910
  • jim_thompson5910
you had the other parts correct
anonymous
  • anonymous
okay then what do i do?
jim_thompson5910
  • jim_thompson5910
so what I'm saying is that f(x) = x^2 + 3x + 2 turns into g(x) = -x^2 - 3x - 2
jim_thompson5910
  • jim_thompson5910
all of the signs flip
anonymous
  • anonymous
and thats the answer?
jim_thompson5910
  • jim_thompson5910
yeah
anonymous
  • anonymous
oh okay thanks
jim_thompson5910
  • jim_thompson5910
np

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