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Graph the equation and its tangent. Graph y = 4x and the tangent to the curve at the point whose x-coordinate is -2. 10

Graph the equation and its tangent. Graph y = 4x and the tangent to the curve at the point whose x-coordinate is -2. 10

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Title: Graphing Quadratic Functions and Tangents** **Graph the Equation and Its Tangent** This exercise involves graphing the quadratic equation and its tangent line at a specific point. **Equation:** Graph \( y = 4x^2 \) and the tangent to the curve at the point whose x-coordinate is \(-2\). **Graph Description:** - The Cartesian coordinate system is used with the x-axis and y-axis intersecting at the origin (0,0). - The x-axis ranges from \(-5\) to \(5\) in increments of 1. - The y-axis ranges from \(-10\) to \(10\) in increments of 5. **Instructions:** 1. Plot the quadratic function \( y = 4x^2 \), which is a parabola opening upwards. Note the vertex is at the origin, and the parabola is symmetric with respect to the y-axis. 2. At \( x = -2 \), calculate the corresponding y-coordinate using the function: \( y = 4(-2)^2 = 16 \). 3. Find the derivative of the function to determine the slope of the tangent line: \( \frac{dy}{dx} = 8x \). 4. Evaluate the derivative at \( x = -2 \): slope = \( 8(-2) = -16 \). 5. Use the point-slope form to derive the equation of the tangent line: \( y - 16 = -16(x + 2) \). This visual and mathematical analysis helps in understanding the behavior of quadratic functions and the concept of tangency in calculus.
Transcribed Image Text:**Title: Graphing Quadratic Functions and Tangents** **Graph the Equation and Its Tangent** This exercise involves graphing the quadratic equation and its tangent line at a specific point. **Equation:** Graph \( y = 4x^2 \) and the tangent to the curve at the point whose x-coordinate is \(-2\). **Graph Description:** - The Cartesian coordinate system is used with the x-axis and y-axis intersecting at the origin (0,0). - The x-axis ranges from \(-5\) to \(5\) in increments of 1. - The y-axis ranges from \(-10\) to \(10\) in increments of 5. **Instructions:** 1. Plot the quadratic function \( y = 4x^2 \), which is a parabola opening upwards. Note the vertex is at the origin, and the parabola is symmetric with respect to the y-axis. 2. At \( x = -2 \), calculate the corresponding y-coordinate using the function: \( y = 4(-2)^2 = 16 \). 3. Find the derivative of the function to determine the slope of the tangent line: \( \frac{dy}{dx} = 8x \). 4. Evaluate the derivative at \( x = -2 \): slope = \( 8(-2) = -16 \). 5. Use the point-slope form to derive the equation of the tangent line: \( y - 16 = -16(x + 2) \). This visual and mathematical analysis helps in understanding the behavior of quadratic functions and the concept of tangency in calculus.
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