Exam Questions - Graph transformations - ExamSolutions - Free Printable

Let's solve the problem step by step.

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Given:

- Curve $ C $: $ y = \frac{2}{x} - 5 $, where $ x \neq 0 $
- Line $ l $: $ y = 4x + 2 $

We are to:

(a) Sketch both graphs on a single diagram and label intercepts.
(b) Find asymptotes of curve $ C $.
(c) Find points of intersection of $ C $ and $ l $.

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## (a) Sketch and Label Graphs

We'll first analyze both curves.

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Curve $ C $: $ y = \frac{2}{x} - 5 $

This is a transformation of the standard hyperbola $ y = \frac{2}{x} $, shifted down by 5 units.

#### Key Features:

1. Vertical Asymptote: $ x = 0 $ (since $ x \neq 0 $)
2. Horizontal Asymptote: As $ x \to \pm\infty $, $ \frac{2}{x} \to 0 $, so $ y \to -5 $
→ So horizontal asymptote: $ y = -5 $
3. Intercepts:
- x-intercept: Set $ y = 0 $
$$
0 = \frac{2}{x} - 5 \Rightarrow \frac{2}{x} = 5 \Rightarrow x = \frac{2}{5}
$$
So x-intercept: $ \left( \frac{2}{5}, 0 \right) $
- y-intercept: Not defined at $ x = 0 $, so no y-intercept

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Line $ l $: $ y = 4x + 2 $

This is a straight line with slope 4 and y-intercept 2.

#### Intercepts:
- y-intercept: When $ x = 0 $, $ y = 2 $ → $ (0, 2) $
- x-intercept: Set $ y = 0 $
$$
0 = 4x + 2 \Rightarrow x = -\frac{1}{2}
$$
→ $ \left( -\frac{1}{2}, 0 \right) $

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Sketch Instructions (for part a):

On a single diagram:
- Draw the hyperbola $ y = \frac{2}{x} - 5 $, which has two branches:
- One in the first quadrant (positive $ x $), approaching $ y = -5 $ as $ x \to \infty $, and going down toward $ -\infty $ as $ x \to 0^+ $
- One in the third quadrant (negative $ x $), approaching $ y = -5 $ as $ x \to -\infty $, and going up toward $ +\infty $ as $ x \to 0^- $
- Draw the line $ y = 4x + 2 $, passing through $ (0, 2) $ and $ (-0.5, 0) $
- Mark:
- $ x $-intercept of $ C $: $ \left( \frac{2}{5}, 0 \right) $
- $ y $-intercept of $ l $: $ (0, 2) $
- $ x $-intercept of $ l $: $ \left( -\frac{1}{2}, 0 \right) $
- Label the asymptotes: $ x = 0 $ (vertical), $ y = -5 $ (horizontal)

> Note: Since this is a text-based explanation, I can't draw the image, but you should sketch accordingly.

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## (b) Asymptotes of Curve $ C $

From earlier analysis:

- Vertical asymptote: $ x = 0 $
- Horizontal asymptote: $ y = -5 $

✔ Answer:
The asymptotes are:
$$
\boxed{x = 0} \quad \text{and} \quad \boxed{y = -5}
$$

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## (c) Points of Intersection of $ y = \frac{2}{x} - 5 $ and $ y = 4x + 2 $

Set the equations equal:
$$
\frac{2}{x} - 5 = 4x + 2
$$

Solve for $ x $:

Bring all terms to one side:
$$
\frac{2}{x} - 5 - 4x - 2 = 0 \Rightarrow \frac{2}{x} - 4x - 7 = 0
$$

Multiply both sides by $ x $ (note: $ x \neq 0 $):
$$
2 - 4x^2 - 7x = 0
\Rightarrow -4x^2 - 7x + 2 = 0
$$

Multiply both sides by $-1$ to make leading coefficient positive:
$$
4x^2 + 7x - 2 = 0
$$

Now solve using quadratic formula:
$$
x = \frac{-7 \pm \sqrt{7^2 - 4(4)(-2)}}{2(4)} = \frac{-7 \pm \sqrt{49 + 32}}{8} = \frac{-7 \pm \sqrt{81}}{8}
$$

$$
\sqrt{81} = 9 \Rightarrow x = \frac{-7 \pm 9}{8}
$$

So:
- $ x = \frac{-7 + 9}{8} = \frac{2}{8} = \frac{1}{4} $
- $ x = \frac{-7 - 9}{8} = \frac{-16}{8} = -2 $

Now find corresponding $ y $ values using $ y = 4x + 2 $:

1. For $ x = \frac{1}{4} $:
$$
y = 4 \cdot \frac{1}{4} + 2 = 1 + 2 = 3
$$
→ Point: $ \left( \frac{1}{4}, 3 \right) $

2. For $ x = -2 $:
$$
y = 4(-2) + 2 = -8 + 2 = -6
$$
→ Point: $ (-2, -6) $

✔ Answer: The points of intersection are:
$$
\boxed{\left( \frac{1}{4}, 3 \right)} \quad \text{and} \quad \boxed{(-2, -6)}
$$

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✔ Final Answers:

(a) Sketch both graphs with:
- Curve $ C $: hyperbola shifted down by 5, with vertical asymptote $ x = 0 $, horizontal asymptote $ y = -5 $, crossing x-axis at $ \left( \frac{2}{5}, 0 \right) $
- Line $ l $: passes through $ (0, 2) $ and $ \left( -\frac{1}{2}, 0 \right) $
- Clearly label these intercepts.

(b) Asymptotes: $ \boxed{x = 0} $ and $ \boxed{y = -5} $

(c) Points of intersection: $ \boxed{\left( \frac{1}{4}, 3 \right)} $ and $ \boxed{(-2, -6)} $

Let me know if you'd like a visual sketch description or help plotting!