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Given that $f(x) = \frac{1}{x}, \quad x \neq 0$, (a) sketch the graph of $y = f(x) + 3$ and state the equations of the asymptotes - Edexcel - A-Level Maths Pure - Question 5 - 2007 - Paper 2

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Given-that---$f(x)-=-\frac{1}{x},-\quad-x-\neq-0$,----(a)-sketch-the-graph-of-$y-=-f(x)-+-3$-and-state-the-equations-of-the-asymptotes-Edexcel-A-Level Maths Pure-Question 5-2007-Paper 2.png

Given that $f(x) = \frac{1}{x}, \quad x \neq 0$, (a) sketch the graph of $y = f(x) + 3$ and state the equations of the asymptotes. (b) Find the coordinates of... show full transcript

Worked Solution & Example Answer:Given that $f(x) = \frac{1}{x}, \quad x \neq 0$, (a) sketch the graph of $y = f(x) + 3$ and state the equations of the asymptotes - Edexcel - A-Level Maths Pure - Question 5 - 2007 - Paper 2

Step 1

(a) sketch the graph of y = f(x) + 3 and state the equations of the asymptotes.

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Answer

To sketch the graph of y=f(x)+3y = f(x) + 3, we first recognize that the base function f(x)=1xf(x) = \frac{1}{x} has vertical asymptote at x=0x = 0 and a horizontal asymptote at y=0y = 0.

With the transformation applied, the graph shifts upward by 3 units. The new horizontal asymptote is now given by the equation:

y=3y = 3

The vertical asymptote remains unchanged at:

x=0x = 0

The overall structure of the graph retains two branches, one in the first quadrant and one in the fourth quadrant, without overlapping the asymptotes.

Step 2

(b) Find the coordinates of the point where y = f(x) + 3 crosses a coordinate axis.

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Answer

To find where the graph crosses the coordinate axes, we need to determine where:

  1. The graph crosses the y-axis:

    • Set x=0x = 0:
      • The graph does not cross the y-axis due to the vertical asymptote.
  2. The graph crosses the x-axis:

    • Set y=0y = 0:

    0=1x+30 = \frac{1}{x} + 3

    • Rearranging gives:
      3=1x-3 = \frac{1}{x}
    • Thus,
      x=13x = -\frac{1}{3}
    • Therefore, the coordinates where y=f(x)+3y = f(x) + 3 crosses the x-axis are:
      (13,0)(-\frac{1}{3}, 0).

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