Question

how to define this equation by switching operations to g-^1 g(x)=-2(x-1)^2 + 7

Answer 1

i solved this didn't i?
http://openstudy.com/users/oldrin.bataku#/updates/4f7259ece4b07738f5ae4d6c

Answer 2

oh yeah but how do we find the x intercepts of that inverse function?

Answer 3

and how come there's a negative sign here => -(x-7)

Answer 4

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

Answer 5

thank you very much

Answer 6

and how do we find the vertex of it?

Answer 7

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)

Answer 8

http://hotmath.com/hotmath_help/topics/vertex-of-a-parabola.html

Answer 9

the vertex is in the exact spot as the original equation but not the inverse function >_>

Answer 10

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

Answer 11

the vertex coordinates swap between g and g^-1 which makes sense because you're swapping the x,y values

Answer 12

ohh i see

Answer 13

but for the x intercept im supposed to also have 2 coordinates is 5,0 one of them?