# Nakafa Framework: LLM

URL: https://nakafa.com/en/exercises/high-school/snbt/quantitative-knowledge/try-out/set-5/14

Exercises: Try Out - Set 5: Real exam simulation to sharpen your skills and build confidence. - Problem 14

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## Exercise 14

### Question

export const metadata = {
  title: "Problem 14",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/23/2025",
};

Given: <InlineMath math="y = -2x^2 - 5x + 7" />, the equation of the tangent line at the point with abscissa <InlineMath math="2" /> is...


### Choices

- [ ] $$-13x - y - 15 = 0$$
- [ ] $$13x - y - 15 = 0$$
- [x] $$13x + y - 15 = 0$$
- [ ] $$-13x + y - 15 = 0$$
- [ ] $$13x + y - 37 = 0$$

### Answer & Explanation

export const metadata = {
  title: "Solution to Problem 14",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/23/2025",
};

Given the quadratic function <InlineMath math="y = -2x^2 - 5x + 7" />. We need to find the equation of the tangent line at the point with abscissa <InlineMath math="x_1 = 2" />.

First, we find the ordinate (<InlineMath math="y_1" />) of the tangent point by substituting <InlineMath math="x_1 = 2" /> into the curve equation:
<MathContainer>
  <BlockMath math="y_1 = -2(2)^2 - 5(2) + 7" />
  <BlockMath math="y_1 = -2(4) - 10 + 7" />
  <BlockMath math="y_1 = -8 - 10 + 7" />
  <BlockMath math="y_1 = -11" />
</MathContainer>
So, the tangent point is <InlineMath math="(2, -11)" />.

Second, we find the gradient of the tangent line (<InlineMath math="m" />) using the first derivative (<InlineMath math="y'" />):
<MathContainer>
  <BlockMath math="y' = -4x - 5" />
</MathContainer>
Substitute <InlineMath math="x = 2" /> into the first derivative to get the gradient:
<MathContainer>
  <BlockMath math="m = -4(2) - 5" />
  <BlockMath math="m = -8 - 5" />
  <BlockMath math="m = -13" />
</MathContainer>

Third, we determine the tangent line equation using the formula <InlineMath math="y - y_1 = m(x - x_1)" />:
<MathContainer>
  <BlockMath math="y - (-11) = -13(x - 2)" />
  <BlockMath math="y + 11 = -13x + 26" />
</MathContainer>
Move all terms to the left side to get the form <InlineMath math="Ax + By + C = 0" />:
<MathContainer>
  <BlockMath math="13x + y + 11 - 26 = 0" />
  <BlockMath math="13x + y - 15 = 0" />
</MathContainer>

So, the equation of the tangent line is <InlineMath math="13x + y - 15 = 0" />.


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