Math Problem Statement

The point

P(5,−2)

lies on the curve

y

2

4− x

.

(a)

If Q is the point

x

2

4− x

,

find the slope of the secant line PQ (correct to six decimal places) for the following values of x.

(i)

4.9

mPQ =

(ii)

4.99

mPQ =

(iii)

4.999

mPQ =

(iv)

4.9999

mPQ =

(v)

5.1

mPQ =

(vi)

5.01

mPQ =

(vii)

5.001

mPQ =

(viii)

5.0001

mPQ =

(b)

Using the results of part (a), guess the value of the slope of the tangent line to the curve at

P(5,−2).

m =

(c)

Using the slope from part (b), find an equation of the tangent line to the curve at

P(5,−2). help me with part c

Solution

Let's work through part (c) step by step.

Part (c) - Finding the Equation of the Tangent Line

You are asked to find the equation of the tangent line to the curve at the point P(5,2)P(5, -2). The slope of the tangent line is found in part (b), where we used the slopes of secant lines to estimate it.

For now, let's assume you have the value of the slope from part (b) (which is the derivative of the function at x=5x = 5).

Let's summarize the general approach:

  1. Tangent Line Formula: The equation of the tangent line to a curve at a given point P(x1,y1)P(x_1, y_1) is given by the point-slope form: yy1=m(xx1)y - y_1 = m(x - x_1) where mm is the slope of the tangent line, and (x1,y1)(x_1, y_1) is the point of tangency.

  2. Given Data:

    • The point PP is (5,2)(5, -2).
    • From part (b), we have the slope mm of the tangent line at x=5x = 5.
    • The equation of the curve is y=24xy = \frac{2}{4 - x}.
  3. Find the Slope of the Tangent Line: The slope mm is the derivative of the function y=24xy = \frac{2}{4 - x} evaluated at x=5x = 5.

Step-by-Step Solution

  1. Find the derivative of the function:

    The function is: y=24xy = \frac{2}{4 - x}

    To differentiate, we use the quotient rule. The quotient rule states that if y=f(x)g(x)y = \frac{f(x)}{g(x)}, then: dydx=f(x)g(x)f(x)g(x)(g(x))2\frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} In our case:

    • f(x)=2f(x) = 2, so f(x)=0f'(x) = 0.
    • g(x)=4xg(x) = 4 - x, so g(x)=1g'(x) = -1.

    Applying the quotient rule: dydx=0(4x)2(1)(4x)2=2(4x)2\frac{dy}{dx} = \frac{0 \cdot (4 - x) - 2 \cdot (-1)}{(4 - x)^2} = \frac{2}{(4 - x)^2}

  2. Evaluate the derivative at x=5x = 5: dydxx=5=2(45)2=2(1)2=2\left.\frac{dy}{dx}\right|_{x=5} = \frac{2}{(4 - 5)^2} = \frac{2}{(-1)^2} = 2

  3. Equation of the Tangent Line: Now, use the point-slope form of the tangent line equation. We know:

    • The point P(5,2)P(5, -2).
    • The slope m=2m = 2.

    The equation of the tangent line is: y(2)=2(x5)y - (-2) = 2(x - 5) Simplifying: y+2=2(x5)y + 2 = 2(x - 5) y+2=2x10y + 2 = 2x - 10 y=2x12y = 2x - 12

Final Answer:

The equation of the tangent line to the curve at P(5,2)P(5, -2) is: y=2x12y = 2x - 12


Would you like me to explain any part of the solution in more detail? Or do you have any further questions on this problem?

Here are 5 related questions you could explore next:

  1. How can we find the second derivative of y=24xy = \frac{2}{4 - x}?
  2. How do we interpret the tangent line's slope in terms of the function's behavior?
  3. What happens to the tangent line when we have a vertical asymptote in the function?
  4. Can we use the tangent line to estimate the value of yy for x=5.1x = 5.1?
  5. What are some applications of tangent lines in real-world problems?

Tip: Always check the units and context when applying a tangent line in real-world scenarios!

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Tangent Lines
Secant Lines
Point-Slope Form

Formulas

Slope of Secant Line: mPQ = (y2 - y1) / (x2 - x1)
Point-Slope Form of a Line: y - y1 = m(x - x1)
Derivative: dy/dx = 2 / (4 - x)^2

Theorems

Quotient Rule for Derivatives
Point-Slope Form of Line Equation

Suitable Grade Level

Grades 11-12