Math Problem Statement
Solution
To find the equation of the tangent line to the curve at , we need to follow these steps:
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Find the derivative of the function to get the slope of the tangent line.
The given equation is . The derivative represents the slope of the tangent line at any point .
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Evaluate the slope at .
Substituting into the derivative:
So, the slope of the tangent line at is 12.
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Find the y-coordinate at .
To get the point where the tangent touches the curve, substitute into the original equation:
Thus, the point of tangency is .
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Use the point-slope form of the equation of a line.
The point-slope form is given by:
where is the slope, and is the point on the line. Substituting , , and :
Simplifying this equation:
So, the equation of the tangent line is .
Would you like any further details or explanations?
Here are some related questions you might find useful:
- How do you find the derivative of more complex functions like trigonometric or exponential functions?
- What is the geometric interpretation of the derivative in terms of slopes and tangents?
- Can the tangent line ever be vertical? Under what conditions?
- How does one find the normal line to a curve at a given point?
- What happens to the tangent line as for polynomials of higher degrees?
Tip: When finding the tangent line, always ensure you evaluate both the slope and the point of tangency correctly for accurate results.
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Math Problem Analysis
Mathematical Concepts
Differentiation
Quadratic Functions
Tangent Lines
Formulas
Derivative of y = 3x^2 - 4 is dy/dx = 6x
Point-slope form of a line: y - y1 = m(x - x1)
Theorems
Derivative provides the slope of the tangent to the curve at any given point.
Suitable Grade Level
Grades 10-12
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