I think most of my answers are right but need confirmation

The directions are to ....

a) Make a sketch of the graph

b) Represent any points of discontinuity with a hole

c) Name domain and range

d) Where appropriate, simplify and name new function

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1. h(x) = sqrt(3x-4)

a) upper half of sideways parabola infirst quad starting at x=4/3

but i dont know what the y-value is

b) no holes?

c) D=[4/3,inf)

R= ?

d) cant be simplified

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2. F(x)= (4x^{2}-1)/(2x+1)

a)line with positive slope with y int at -1

b) hole at (-1/2,-2)

c) D=(-inf,-1/2),(-1/2,inf)

R=(-inf, -2), (-2,inf)

d) new function = 2x-1

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3. f(x)=(x^{2}-4x+3)/(x-1)

a) line with positive slope with y-int at -3

b) hole at (1,-2)

c) D=(-inf, 1),(1,inf)

R=(-inf,-2),(-2,inf)

d) new function = x-3

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4. This ones a piecewise function

f(x)={2x-1 if x cant equal 2

{0 if x=2

a)line with y int at -1 and point at (2,0)

b) hole at (2,3)

c) D=all reals

R=(-inf,3), (3,inf)

d)cant be simplified

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5. also a piecewise function

h(x)={x+6 if x is less than or equal to -4

{sqrt(16-x^{2}) if -4<x<4

{6-x if x is greater than or equal to 4

a) semi circle in top half of graph with ends at 4 and -4 but open circles, one line with negative slope in second quad above x-axis at -4 and the same image flipped in other quad

b) holes at (-4,0) and (4,0)

c) D= all reals

R=(-inf, 4)

d) cant be simplified

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6. also piecewise function

g(x)={x-2 if x<0

{0 if x=0

{x^{2}+1 if x>0

a)right side of parabola with open circle on y int, point at the origin, line with negative slope y-int at -2 as open circle

b)holes at (0,1) and (-2,0)

c) D=all reals

R=(-inf,-2),[2,2],(1,inf)

d) cant be simplified

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7. F(x)= (x^{3}+3x^{2}+x+3)/(x+3)

a) parabola centered at origgin with hole at x-value of -3

b) hole at (-3,10)

c) D=(-inf, inf),(x cant equal -3)

R= [0,inf), (y cant equal...10?)

d) new function = x^{2}+1

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Thanks