Can someone help me with this and show me how to do it?

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  5i) f(x) = 3·13^x +5

  5ii) f(x) = -6·(1/2)^x +5

  6) f(x) = 3·8^x -1

  9a) (1, 0), (0, -3)

  9b) (2, 0), (0, 8)

Step-by-step explanation:

5. The horizontal asymptote is y = c. To meet the requirements of the problem, you must choose c=5 and any other (non-zero) numbers for 'a' and 'b'. (You probably want 'b' to be positive, so as to avoid complex numbers.)

i) f(x) = 3·13^x +5

ii) f(x) = -6·(1/2)^x +5

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6. You already know c=-1, so put x=0 in the equation and solve for 'a'. As in problem 5, 'b' can be any positive value.

  f(0) = 2 = a·b^0 -1

  3 = a

One possible function is ...

  f(x) = 3·8^x -1

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9. The x-intercept is the value of x that makes y=0. We can solve for the general case:

  0 = a·b^x +c

  -c = a·b^x

  -c/a = b^x

Taking logarithms, we have ...

  log(-c/a) = x·log(b)

Of course, the y-intercept is (a+c), since the b-factor is 1 when x=0.

a) x-intercept: log2(6/3) = log2(2) = 1, or point (1, 0)

   y-intercept: 3-6 = -3, or point (0, -3)

b) x-intercept: log3(9/1) = log3(3^2) = 2, or point (2, 0)

  y-intercept: -1 +9 = 8, or point (0, 8)

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Additional comment

It is nice to be comfortable with logarithms. It can be helpful to remember that a logarithm is an exponent. Even so, you can solve the x-intercepts of problem 9 using the expression we had just before taking logarithms.

  a) 6/3 = 2^x   ⇒   2^1 = 2^x   ⇒   x=1

  b) -9/-1 = 3^x   ⇒   3^2 = 3^x   ⇒   x=2

y=mx+b

step-by-step explanation:

m slope, b y-intercept

isaiah walks 3 miles due north, turns and then 5 miles due east. how far is he from his starting point?

a^2 + b^2 = c^2

let c = distance from starting point.

(3)^2 + (5)^2 = c^2

9 + 25 = c^2

34 = c^2

sqrt{34} = sqrt{c^2}

sqrt{34} = c