Question

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

Answer 1

9514 1404 393

Answer:

  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)

  $\displaystyle x=\frac{\log\left(-\dfrac{c}{a}\right)}{\log(b)}=\log_b\left(-\dfrac{c}{a}\right)$

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