SOLUTION: Please help me solve this problem: Show that the expression for the slope of the secant line through y=x^2 +3x at x=3 and x=3+h is msec=9+h. Sketch the graph of y=x^2 +3x sh

Algebra ->  Test -> SOLUTION: Please help me solve this problem: Show that the expression for the slope of the secant line through y=x^2 +3x at x=3 and x=3+h is msec=9+h. Sketch the graph of y=x^2 +3x sh      Log On


   

Question 1005325: Please help me solve this problem:
Show that the expression for the slope of the secant line through y=x^2 +3x at x=3 and x=3+h is msec=9+h.
Sketch the graph of y=x^2 +3x showing a sequence of secant lines described in the previous question as h approaches 0.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The function is y=x%5E2+%2B3x .
The secant requested is the straight line that passes through points
A with system%28x%5BA%5D=3%2Cy%5BA%5D=3%5E2%2B3%2A3=9%2B9=18%29 and
B with
The slope of secant AB is
m=%28y%5BB%5D-y%5BA%5D%29%2F%28x%5BB%5D-x%5BA%5D%29 --> m=%2818%2B9h%2Bh%5E2-18%29%2F%283%2Bh-3%29 --> m=%289h%2Bh%5E2%29%2Fh --> m=h%289%2Bh%29%2Fh --> highlight%28m=9%2Bh%29

The graph with the secants for h=1, h=2, and h=3 are shown below.