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Given: f(x) = -1/(x-3) + 2. Write down the equations of the asymptotes of f, the domain of f, determine the coordinates of the x-intercept of f, write down the coordinates of the y-intercept of f, and draw the graph of f.

Mathematics

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4 parts
5.1
3
2
5.2

x∈R,x≠3x \in \mathbb{R}, x \neq 3

5.3

(72,0)\left(\frac{7}{2}, 0\right)

5.4

(0,73)\left(0, \frac{7}{3}\right)

Graph of f(x) = −1/(x − 3) + 2

Use the left and right arrow keys to trace the curve, up and down to jump between key points, and plus and minus to zoom.

Key points of f(x) = −1/(x − 3) + 2
Pointxy
y-intercept02.33
x-intercept3.50

Step-by-step solution

8 steps

Part 5.1· 2 steps

Write down the equations of the asymptotes of ff.

  1. Identify the form of the function.

    The function is given as f(x)=−1x−3+2f(x) = \frac{-1}{x-3} + 2, which is in the form f(x)=ax−h+kf(x) = \frac{a}{x-h} + k.

  2. Determine the vertical and horizontal asymptotes.

    The vertical asymptote is x=hx=h and the horizontal asymptote is y=ky=k.
    From the given function, h=3h=3 and k=2k=2.
    The equations of the asymptotes are:

    x=3x = \boxed{3} y=2y = \boxed{2}

Part 5.2

x∈R,x≠3x \in \mathbb{R}, x \neq 3

Write down the domain of ff.

  1. Identify values for which the denominator is zero.

    The denominator of the function is x−3x-3. The function is undefined when x−3=0x-3=0.

  2. Solve for xx.

    x−3=0  ⟹  x=3x-3=0 \implies x=3.
    The domain is all real numbers except where the function is undefined.
    The domain of ff is x∈R,x≠3\boxed{x \in \mathbb{R}, x \neq 3}.

Part 5.3

(72,0)\left(\frac{7}{2}, 0\right)

Determine the coordinates of the xx-intercept of ff.

  1. Set f(x)=0f(x)=0 to find the xx-intercept.

    0=−1x−3+20 = \frac{-1}{x-3} + 2

  2. Solve for xx.

    −2=−1x−3-2 = \frac{-1}{x-3} 2=1x−32 = \frac{1}{x-3} 2(x−3)=12(x-3) = 1 2x−6=12x - 6 = 1 2x=72x = 7 x=72x = \frac{7}{2}

    The coordinates of the xx-intercept are (72,0)\boxed{\left(\frac{7}{2}, 0\right)}.

Part 5.4

(0,73)\left(0, \frac{7}{3}\right)

Write down the coordinates of the yy-intercept of ff.

  1. Set x=0x=0 to find the yy-intercept.

    f(0)=−10−3+2f(0) = \frac{-1}{0-3} + 2

  2. Calculate f(0)f(0).

    f(0)=−1−3+2f(0) = \frac{-1}{-3} + 2 f(0)=13+2f(0) = \frac{1}{3} + 2 f(0)=13+63f(0) = \frac{1}{3} + \frac{6}{3} f(0)=73f(0) = \frac{7}{3}

    The coordinates of the yy-intercept are (0,73)\boxed{\left(0, \frac{7}{3}\right)}.

Part 5.5

Draw the graph of ff. Clearly show ALL the asymptotes and intercepts with the axes.

To draw the graph:

  • Draw the vertical asymptote at x=3x=3 as a dashed line.
  • Draw the horizontal asymptote at y=2y=2 as a dashed line.
  • Plot the xx-intercept at (72,0)(\frac{7}{2}, 0) or (3.5,0)(3.5, 0).
  • Plot the yy-intercept at (0,73)(0, \frac{7}{3}) or (0,≈2.33)(0, \approx 2.33).
  • Since the numerator is negative (−1-1), the branches of the hyperbola will be in the second and fourth quadrants relative to the asymptotes. Sketch the two branches of the hyperbola approaching the asymptotes and passing through the intercepts.
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