11.4. Solved Problems

1. Change to slope-intercept form: 3x - 2y = -6

   -2y = -3x - 6
   
   y = 1.5x + 3
   

2. Change to slope-intercept form: 12x + 3y = +6

   3y = -12x + 6
   
   y = -4x + 2
   

3. Find the equation of a line in the form Ax + B + C = 0 for a line with a slope of -2 and a point at {4, 6}.

   -2 = (y - 6)/(x - 4)
   
   -2(x - 4) = y - 6
   
   -2x + 8 = y - 6
   
   -2x - y = -14
   
   2x + y - 14 = 0
   

4. Find the equation of a line in the form Ax + By + C = 0 that includes the points {2, 3} and {-5, 8}.

     8 - 3     y - 3
   -------- = -------
    -5 - 2     x - 2
   
     5     y - 3
   ---- = -------
    -7     x - 2
   
   5(x - 2) = -7(y - 3)
   
   5x - 10 = -7y + 21
   
   5x + 7y - 31 = 0
   

5. Find the equation of the line with a slope of 5/2 and a y-intercept 2.

    5     y - 2
   --- = -------
    2     x - 0
   
   5x = 2y - 4
   
   5x - 2y + 4 = 0
   

6. Find the equation of a line with x-intercept of 3 and slope of -2.

         y - 0
   -2 = -------
         x - 3
   
   -2x + 6 = y
   
   2x + y - 6 = 0
   

7. Find the equation of a line parallel to y = 3x and passing through the point {1, 1}.

   3 = (y - 1)/(x - 1)
   
   3x - 3 = y - 1
   
   3x - y - 2 = 0
   

8. Find the equation of a line perpendicular to y = 0.5x + 2 and passing through the point {4, 4}.

   2 = (y - 4)/(x - 4)
   
   2x - 8 = y - 4
   
   2x - y - 4 = 0
   

9. It is known from calculus that a line tangent to the curve f(x) = x² + 2x + 1 has a slope of 2x + 2 at the point {x, f(x)}. Find the equation of the line tangent to f(x) at x = 1.

   y = f(1) = 12 + 2(1) + 1 = 4, point of tangency is {1, 4}
   
   m = 2(1) + 2 = 4
   
   4 = (y - 4)/(x - 1)
   
   4x - 4 = y - 4
   
   4x - y = 0
   

10. Describe end behavior for: f(x) = x⁴ + 7x³ + 2x - 9

    f(x) --> +∞ as x --> -∞ and 
    f(x) --> +∞ as x --> +∞ 
    

11. Describe end behavior for: f(x) = -x⁵ + 3x⁴ - 7x² + 2x + 1

    f(x) --> +∞ as x --> -∞ and 
    f(x) --> -∞ as x --> +∞ 
    

12. Describe end behavior for: f(x) = x² - 2x + 8

    f(x) --> +∞ as x --> -∞ and 
    f(x) --> +∞ as x --> +∞ 
    

13. Describe end behavior for: f(x) = -x⁴ - 8x³ + 4x

    f(x) --> -∞ as x --> -∞ and 
    f(x) --> -∞ as x --> +∞ 
    

14. Describe end behavior for: f(x) = x³ + 2x² + 2x - 1

    f(x) --> -∞ as x --> -∞ and 
    f(x) --> +∞ as x --> +∞ 
    

15. Find the zeroes and extrema for the function: f(x) = x² + x + 1

    Use the quadratic formula to find zeroes,
               _________            ___
         -b ± √ b2 - 4ac      -1 ± √ -3 
    x = ---------------- = --------------- 
               2a                 2
    
    Then, zeroes are complex (graph does not cross x-axis).
    
    g(x) = 2x + 1 = 0 --> critical point at x = -0.5
    h(x) = 2          --> concave up 
    The graph has minimum at x = -0.5
    

16. Find the zeroes and extrema for the function: f(x) = x⁴ + 2x² + 1

    f(x) = (x2 + 1)(x2 + 1) 
         = (x + i)(x - i)(x + i)(x - i)
    has only complex root, no zeroes.
    
    g(x) = 4x3 + 4x = 4x(x2 + 1)
    The second factor has complex roots.  
    So there is one critical point at x = 0
    
    h(0) = 12(0)2 + 4 = 4
    The function has a minimum at {0, 1}
    

17. Find the zeroes and extrema for the function: f(x) = x³ - 2x² + 3x - 2

    f(x) = (x - 1)(x2 - x + 2)
    The determinant of the second factor is: 12 - 4(1)(2) = -7
    So it has complex roots.  There is one zero at x = 1
    
    g(x) = 3x2 - 4x + 3
    The determinant of g(x) is: 42 - 4(3)(3) = -20
    So there are no critical points.
    

18. Find the zeroes and extrema for the function: f(x) = x³ + 2x² - 5x - 6

    f(x) = (x + 3)(x + 1)(x - 2)
    The zeroes are at x = -3, -1, 2
    
    g(x) = 3x2 + 4x - 5
    The critical points are at x = -2.12, 0.79
    
    h(x) = 6x + 4
    h(-2.12) = -8.7
    h(+0.79) = +8.7
    
    Maximum at: {-2.12, +4.06}
    Minimum at: {+0.79, -8.21}
    

19. Graph the function: f(x) = x³ + x² - x - 1

    

20. Graph the function: f(x) = x⁴ - 2x² + 1