NSC Mathematics Functions: Consider: g(x) = \frac{-a}{x +

Consider: g(x) = \frac{-a}{x + 2} + q The following information of g is given: - Domain: x ∈ ℝ; x ≠ -2 - x-intercept at K(1 ; 0) - y-intercept at N(0 ; -\frac{1}{2}) 5.1 Show that the equation of g is given by: g(x) = \frac{-3}{x + 2} + 1 5.2 Write down the range of g - NSC Mathematics - Question 5 - 2022 - Paper 1

Question 5

$Consider:---g(x)-=-\frac{-a}{x-+-2}-+-q----The-following-information-of-g-is-given:-----Domain:-x-∈-ℝ;-x-≠--2-----x-intercept-at-K(1-;-0)-----y-intercept-at-N(0-;--\frac{1}{2})----5.1-Show-that-the-equation-of-g-is-given-by:-g(x)-=-\frac{-3}{x-+-2}-+-1----5.2-Write-down-the-range-of-g-NSC Mathematics-Question 5-2022-Paper 1.png$

Consider: g(x) = \frac{-a}{x + 2} + q The following information of g is given: - Domain: x ∈ ℝ; x ≠ -2 - x-intercept at K(1 ; 0) - y-intercept at N(0 ; -\... show full transcript

Worked Solution & Example Answer:Consider: g(x) = \frac{-a}{x + 2} + q The following information of g is given: - Domain: x ∈ ℝ; x ≠ -2 - x-intercept at K(1 ; 0) - y-intercept at N(0 ; -\frac{1}{2}) 5.1 Show that the equation of g is given by: g(x) = \frac{-3}{x + 2} + 1 5.2 Write down the range of g - NSC Mathematics - Question 5 - 2022 - Paper 1

Step 1

5.1 Show that the equation of g is given by: g(x) = \frac{-3}{x + 2} + 1

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Answer

To show that the given equation for g is valid, we can substitute the x-intercept K(1; 0) into the equation:

Substituting into the function:

g(1) = \frac{-a}{1 + 2} + q = 0$$ This simplifies to:

0 = \frac{-a}{3} + q

Rearranging gives us:

a = 3q

Next, substituting the y-intercept N(0; -\frac{1}{2}) into the equation:

g(0) = \frac{-a}{0 + 2} + q = -\frac{1}{2}$$

This simplifies to:

-\frac{a}{2} + q = -\frac{1}{2}\

Substituting q from earlier:

-\frac{a}{2} + \frac{a}{3} = -\frac{1}{2} \

Multiply through by 6 to eliminate the denominators:

-3a + 2a = -3\

This simplifies to:

a = -3\

Thus, we can conclude that the equation simplifies to:

g(x) = \frac{-3}{x + 2} + 1$$

Step 2

5.2 Write down the range of g.

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Answer

The range of g can be deduced from its vertical asymptote. Since g approaches ±∞ as x approaches -2, and its maximum value does not reach 1, the range is:

y \in (-\infty, 1) \cup (1, +\infty)$$

Step 3

5.3 Determine the equation of h, the axis of symmetry of g, in the form y = mx + c, where m > 0.

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Answer

To find the axis of symmetry h, we determine that g is a rational function which generally has symmetry about its vertical asymptote. The vertical asymptote occurs at x = -2. The axis of symmetry can be described as:

h: y = -1 \text{ (the horizontal line at y = -1)}$$

Step 4

5.4 Write down the coordinates of K', the image of K reflected over h.

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Answer

To find the coordinates of K', which is the reflection of K(1; 0) over the line y = -1, we first find the distance from K to the line:

$ext{Distance} = 0 - (-1) = 1$

Thus, the reflected point K' will be:

$K' = (1, -1 - 1) = (1, -2)$

5.1 Show that the equation of g is given by: g(x) = \frac{-3}{x + 2} + 1

5.2 Write down the range of g.

5.3 Determine the equation of h, the axis of symmetry of g, in the form y = mx + c, where m > 0.

5.4 Write down the coordinates of K', the image of K reflected over h.

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