# Nakafa Framework: LLM

URL: https://nakafa.com/en/exercises/high-school/snbt/quantitative-knowledge/try-out/set-7/2

Exercises: Try Out - Set 7: Real exam simulation to sharpen your skills and build confidence. - Question 2

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## Exercise 2

### Question

export const metadata = {
  title: "Question 2",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/24/2025",
};

Given <InlineMath math="y = -x^2 - 2x + 8" />, the equation of the tangent line at the point with abscissa <InlineMath math="-2" /> is...


### Choices

- [ ] $$y = 2x - 4$$
- [ ] $$y = -2x - 2$$
- [ ] $$y = -2x + 4$$
- [x] $$y = 2x + 12$$
- [ ] $$y = -2x + 12$$

### Answer & Explanation

export const metadata = {
  title: "Question 2 Explanation",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/24/2025",
};

Given the function <InlineMath math="y = -x^2 - 2x + 8" />. We want to find the equation of the tangent line at the point with abscissa <InlineMath math="x = -2" />.

#### Finding the Ordinate

Substitute <InlineMath math="x = -2" /> into the curve equation:

<MathContainer>
  <BlockMath math="y = -(-2)^2 - 2(-2) + 8" />
  <BlockMath math="y = -4 + 4 + 8" />
  <BlockMath math="y = 8" />
</MathContainer>

So, the point of tangency is <InlineMath math="(-2, 8)" />.

#### Finding the Gradient of the Tangent Line

The gradient of the tangent line <InlineMath math="m" /> is the value of the first derivative <InlineMath math="y'" /> at the abscissa of the point of tangency.

<MathContainer>
  <BlockMath math="y' = -2x - 2" />
</MathContainer>

Substitute <InlineMath math="x = -2" />:

<MathContainer>
  <BlockMath math="m = -2(-2) - 2" />
  <BlockMath math="m = 4 - 2" />
  <BlockMath math="m = 2" />
</MathContainer>

#### Determining the Equation of the Tangent Line

Use the point-slope form for the equation of a line passing through <InlineMath math="(x_1, y_1)" /> with gradient <InlineMath math="m" />:

<MathContainer>
  <BlockMath math="y - y_1 = m(x - x_1)" />
</MathContainer>

Substitute <InlineMath math="x_1 = -2" />, <InlineMath math="y_1 = 8" />, and <InlineMath math="m = 2" />:

<MathContainer>
  <BlockMath math="y - 8 = 2(x - (-2))" />
  <BlockMath math="y - 8 = 2(x + 2)" />
  <BlockMath math="y - 8 = 2x + 4" />
  <BlockMath math="y = 2x + 4 + 8" />
  <BlockMath math="y = 2x + 12" />
</MathContainer>

Thus, the equation of the tangent line is <InlineMath math="y = 2x + 12" />.


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