What are the domain,range and asymptote of h(x)=6^x-4
2 answers:
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Answer:
y=9x+20
Step-by-step explanation:
5x-12y-17=0 (plus 17 from both sides)
5x-12y=17 (plus 12y from both sides)
5x=12y+17 (divide 5 from everything)
5x-12y-17=0 (plus 17 from both sides)
5x-12y=17 (minus 5x from both sides)
-12y=-5x+17 (divide everything by -12)
1*9(x+2)-(y+1)+3=0 (distribute 9 to x and to 2) (distribute -1 to y and to 1)
1*9x+18-y-1+3=0 (times 9x by 1)
9x+18-y-1+3=0 (combine like terms)
9x-y+20=0 (minus 20 from both sides)
9x-y=-20 (plus y to both sides)
9x=y-20 (divide 9 from both sides)
1*9(x+2)-(y+1)+3=0 (distribute 9 to x and to 2) (distribute -1 to y and to 1)
1*9x+18-y-1+3=0 (times 9x by 1)
9x+18-y-1+3=0 (combine like terms)
9x-y+20=0 (minus 20 from both sides)
9x-y=-20 (minus 9x from both sides)
-y=-9x-20 (divide -1 from everything)
y=9x+20
9514 1404 393
Answer:
a (3,-1)
Step-by-step explanation:
The number that "completes the square" is the square of half the x-coefficient, (-6/2)^2 = 9. Rearranging the given function to include the square trinomial, we have ...
f(x) = x^2 -6x +9 -1 . . . . . . . here, we have 8 = 9 - 1
f(x) = (x -3)^2 -1 . . . . . . . . . . vertex form
Comparing this to the generic vertex form ...
f(x) = (x -h)^2 +k . . . . . . . vertex at (h, k)
we see that h=3 and k=-1.
The vertex is (h, k) = (3, -1).
