Math Problem Statement

Given the function f, of, x, equals, x, cubed, plus, 6, x, squared, plus, 7, x, minus, 2f(x)=x 3 +6x 2 +7x−2, determine all coordinate points left parenthesis, x, comma, y, right parenthesis(x,y) on the graph of ff such that the line tangent to ff at left parenthesis, x, comma, y, right parenthesis(x,y) has a slope of minus, 2, .−2

Solution

To determine all coordinate points (x,y)(x, y) on the graph of the function f(x)=x3+6x2+7x2f(x) = x^3 + 6x^2 + 7x - 2 where the tangent line has a slope of 2-2, we need to follow these steps:

Step 1: Find the derivative of f(x)f(x)

The derivative of a function gives the slope of the tangent line at any point xx. So, let's differentiate f(x)f(x):

f(x)=ddx(x3+6x2+7x2)f'(x) = \frac{d}{dx}(x^3 + 6x^2 + 7x - 2)

Using the power rule, we get:

f(x)=3x2+12x+7f'(x) = 3x^2 + 12x + 7

Step 2: Set the derivative equal to the slope 2-2

Since we are looking for the points where the slope of the tangent line is 2-2, set the derivative equal to 2-2:

3x2+12x+7=23x^2 + 12x + 7 = -2

Step 3: Solve for xx

Rearrange the equation to set it equal to zero:

3x2+12x+7+2=03x^2 + 12x + 7 + 2 = 0

3x2+12x+9=03x^2 + 12x + 9 = 0

Now, divide the whole equation by 3:

x2+4x+3=0x^2 + 4x + 3 = 0

This is a quadratic equation. We can factor it as:

(x+3)(x+1)=0(x + 3)(x + 1) = 0

Thus, the solutions for xx are:

x=3orx=1x = -3 \quad \text{or} \quad x = -1

Step 4: Find the corresponding yy-coordinates

To find the corresponding yy-coordinates, substitute x=3x = -3 and x=1x = -1 into the original function f(x)f(x):

  • For x=3x = -3:

f(3)=(3)3+6(3)2+7(3)2=27+54212=4f(-3) = (-3)^3 + 6(-3)^2 + 7(-3) - 2 = -27 + 54 - 21 - 2 = 4

So, the point is (3,4)(-3, 4).

  • For x=1x = -1:

f(1)=(1)3+6(1)2+7(1)2=1+672=4f(-1) = (-1)^3 + 6(-1)^2 + 7(-1) - 2 = -1 + 6 - 7 - 2 = -4

So, the point is (1,4)(-1, -4).

Step 5: Final answer

The two coordinate points where the slope of the tangent line is 2-2 are:

(3,4)and(1,4)(-3, 4) \quad \text{and} \quad (-1, -4)

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

  1. How do you find the tangent line equation at a specific point on a curve?
  2. What is the geometric meaning of the derivative of a function at a point?
  3. How does the concavity of a function affect the shape of its graph?
  4. How can you use derivatives to find local maxima and minima of a function?
  5. What are the steps to solving cubic equations for their roots?

Tip: Remember, the derivative gives the instantaneous rate of change (or slope) of the function at any given point!

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Math Problem Analysis

Mathematical Concepts

Differentiation
Quadratic Equations
Tangent Lines

Formulas

f'(x) = 3x^2 + 12x + 7
Quadratic equation: ax^2 + bx + c = 0

Theorems

Power rule for differentiation
Quadratic formula for solving ax^2 + bx + c = 0

Suitable Grade Level

Grades 10-12