question 10 (4 points)

determine an equation of the line tangent to the curve $y = \sqrt{x - 8}$ at the point with $x$-coordinate 9.

$x-2y+8=0$
$-x+2y+8=0$
$x-2y+7=0$
$-x+2y+7=0$

Question

question 10 (4 points)

determine an equation of the line tangent to the curve $y = \sqrt{x - 8}$ at the point with $x$-coordinate 9.

$x-2y+8=0$
$-x+2y+8=0$
$x-2y+7=0$
$-x+2y+7=0$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Linear Equations", "First Principles Derivative" ], "new_concepts": [ "Tangent Line Equation", "Derivative of Radical Functions" ], "current_concepts": [ "Linear Equations", "First Principles Derivative", "Tangent Line Equation", "Derivative of Radical Functions" ] } ## Find the point of tangency Using the **Linear Equations** knowledge point $$ \begin{aligned} &x = 9\ &y = \sqrt{9 - 8} = \sqrt{1} = 1\ &\text{Point of tangency: } (9, 1) \end{aligned} $$ ## Find the derivative of the curve Using the **First Principles Derivative** knowledge point $$ \begin{aligned} &y' = \frac{d}{dx}(x - 8)^{1/2} = \frac{1}{2\sqrt{x - 8}} \end{aligned} $$ ## Calculate the slope of the tangent line Evaluate the derivative at $x = 9$: $$ \begin{aligned} &m = y'(9) = \frac{1}{2\sqrt{9 - 8}} = \frac{1}{2} \end{aligned} $$ ## Write the equation of the tangent line Using the **Linear Equations** knowledge point $$ \begin{aligned} &y - 1 = \frac{1}{2}(x - 9)\ &2y - 2 = x - 9\ &x - 2y - 7 = 0 \quad \text{or} \quad -x + 2y + 7 = 0 \end{aligned} $$

x-2y+8=0

-x+2y+8=0

x-2y+7=0

-x+2y+7=0

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Calculus", "Tangent Line Equation" ] }

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Find the point of tangency

Find the derivative of the curve

Calculate the slope of the tangent line

Write the equation of the tangent line

Answer: