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Identify the y-intercept.

Mathematics

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6 parts
5.1

(0,1)\text{} (0, 1) \text{}

5.2

23\text{}\frac{2}{3}\text{}

5.3

h(x)=(32)x\text{}h(x) = \left(\frac{3}{2}\right)^x\text{}

5.4

y=log⁡32x\text{}y = \log_{\frac{3}{2}} x\text{}

5.5

0<x≤32\text{}0 < x \leq \frac{3}{2}\text{}

5.6

x>3\text{}x > 3\text{}

Step-by-step solution

13 steps

Part 5.1· 1 step

Write down the coordinates of B.

  1. Identify the y-intercept.

    B is the y-intercept of the function g(x)=axg(x) = a^x. To find the y-intercept, set x=0x=0.

    g(0)=a0=1g(0) = a^0 = 1

    The coordinates of B are (0,1)\boxed{\text{} (0, 1) \text{}}

Part 5.2

23\text{}\frac{2}{3}\text{}

Determine the value of a.

  1. Substitute the coordinates of point A into the function.

    Point A (−3,278)(-3, \frac{27}{8}) lies on g(x)=axg(x) = a^x.

    278=a−3\frac{27}{8} = a^{-3}
  2. Solve for aa.

    278=1a3\frac{27}{8} = \frac{1}{a^3} a3=827a^3 = \frac{8}{27}

    Take the cube root of both sides:

    a=8273a = \sqrt[3]{\frac{8}{27}} a=23a = \frac{2}{3}

    The value of aa is 23\boxed{\text{}\frac{2}{3}\text{}}

Part 5.3

h(x)=(32)x\text{}h(x) = \left(\frac{3}{2}\right)^x\text{}

Determine the equation of h if h is the reflection of g in the y-axis.

  1. Apply the reflection rule.

    A reflection in the y-axis means replacing xx with −x-x in the function.
    Given g(x)=(23)xg(x) = (\frac{2}{3})^x.

    h(x)=g(−x)h(x) = g(-x) h(x)=(23)−xh(x) = \left(\frac{2}{3}\right)^{-x}
  2. Simplify the expression.

    h(x)=(32)xh(x) = \left(\frac{3}{2}\right)^x

    The equation of hh is h(x)=(32)x\boxed{\text{}h(x) = \left(\frac{3}{2}\right)^x\text{}}

Part 5.4

y=log⁡32x\text{}y = \log_{\frac{3}{2}} x\text{}

Write down the equation of h−1(x)h^{-1}(x), the inverse of h, in the form y=…y = \dots.

  1. Start with the equation for h(x)h(x).

    y=(32)xy = \left(\frac{3}{2}\right)^x
  2. Swap xx and yy to find the inverse.

    x=(32)yx = \left(\frac{3}{2}\right)^y
  3. Solve for yy by converting to logarithmic form.

    y=log⁡32xy = \log_{\frac{3}{2}} x

    The equation of h−1(x)h^{-1}(x) is y=log⁡32x\boxed{\text{}y = \log_{\frac{3}{2}} x\text{}}

Part 5.5

0<x≤32\text{}0 < x \leq \frac{3}{2}\text{}

For which values of x will h−1(x)≤1h^{-1}(x) \leq 1?

  1. Set up the inequality using the inverse function.

    log⁡32x≤1\log_{\frac{3}{2}} x \leq 1
  2. Convert the logarithmic inequality to an exponential inequality.

    Since the base b=32b = \frac{3}{2} is greater than 1, the inequality direction remains the same.

    x≤(32)1x \leq \left(\frac{3}{2}\right)^1 x≤32x \leq \frac{3}{2}
  3. Consider the domain of the logarithmic function.

    The argument of a logarithm must be positive.

    x>0x > 0

    Combining the conditions, the values of xx are 0<x≤32\boxed{\text{}0 < x \leq \frac{3}{2}\text{}}

Part 5.6

x>3\text{}x > 3\text{}

Write down the domain of h−1(x−3)h^{-1}(x - 3).

  1. Recall the domain of h−1(x)h^{-1}(x).

    The domain of h−1(x)=log⁡32xh^{-1}(x) = \log_{\frac{3}{2}} x is x>0x > 0.

  2. Apply the domain condition to h−1(x−3)h^{-1}(x - 3).

    For h−1(x−3)h^{-1}(x - 3), the argument of the logarithm is (x−3)(x - 3). This argument must be greater than 0.

    x−3>0x - 3 > 0 x>3x > 3

    The domain of h−1(x−3)h^{-1}(x - 3) is x>3\boxed{\text{}x > 3\text{}}

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