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Identify the graph as a hyperbola, determine its equation, domain, range, and whether it is a function, given its asymptotes and a point.

Mathematics

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3 parts
2.2
1

g(x)=1x+1g(x) = \frac{1}{x} + 1

2.4

x∈R,x≠0x \in \mathbb{R}, x \ne 0

y∈R,y≠1y \in \mathbb{R}, y \ne 1

2.6

y=x+1y = x + 1

y=−x+1y = -x + 1

Step-by-step solution

4 steps

Here are the solutions to the questions:

Step 1: Identify the type of graph and asymptotes.
The given graph is a hyperbola. The general form of the function is g(x)=ax+qg(x) = \frac{a}{x} + q.
From the graph, the vertical asymptote is the y-axis, x=0x=0.
The horizontal asymptote is the line y=1y=1.

Part 2.1

The mathematical name of this graph is a hyperbola (or a reciprocal function).

Part 2.21

The horizontal asymptote is y=qy=q. From the graph, the horizontal asymptote is y=1y=1.

Therefore, the value of qq is 1\boxed{1}.

  1. Determine the equation of the graph.

    We know g(x)=ax+qg(x) = \frac{a}{x} + q. From 2.2, q=1q=1, so g(x)=ax+1g(x) = \frac{a}{x} + 1.
    To find aa, we can use a point from the graph. Let's use the point (1,2)(1, 2) which is clearly visible on the graph.
    Substitute x=1x=1 and g(x)=2g(x)=2 into the equation:

    2=a1+12 = \frac{a}{1} + 1 2=a+12 = a + 1 a=2−1a = 2 - 1 a=1a = 1

    Therefore, the equation of the graph is g(x)=1x+1\boxed{g(x) = \frac{1}{x} + 1}.

  2. Determine the domain and range.

Part 2.4

x∈R,x≠0x \in \mathbb{R}, x \ne 0

The domain of g(x)g(x) is all real numbers except where the denominator is zero, which is x=0x=0.

The range of g(x)g(x) is all real numbers except the horizontal asymptote, which is y=1y=1.
Domain: x∈R,x≠0\boxed{x \in \mathbb{R}, x \ne 0}
Range: y∈R,y≠1\boxed{y \in \mathbb{R}, y \ne 1}

  1. Determine if the graph is a function and provide a reason.

Part 2.5· 1 step

The graph is a *function

Reason: For every input value of xx (except x=0x=0), there is exactly one output value of yy. This can be verified by the vertical line test, where any vertical line drawn on the graph intersects the graph at most once.

  1. Write down the two equations of symmetry.

Part 2.6

y=x+1y = x + 1

For a hyperbola of the form y=ax+qy = \frac{a}{x} + q, the lines of symmetry pass through the intersection of the asymptotes (0,q)(0, q) and have slopes of 1 and −1-1.

The intersection of the asymptotes is (0,1)(0, 1).
The first line of symmetry has a slope of 1:

y−y1=m(x−x1)y - y_1 = m(x - x_1) y−1=1(x−0)y - 1 = 1(x - 0) y=x+1\boxed{y = x + 1}

The second line of symmetry has a slope of −1-1:

y−y1=m(x−x1)y - y_1 = m(x - x_1) y−1=−1(x−0)y - 1 = -1(x - 0) y=−x+1\boxed{y = -x + 1}
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