Question
Please show work, thanks!
slopes of lines are easy to calculate using a difference quotient, but you need Two points an the line, and only have one. If we calculate the slope of the line between two points on a curve, we get the average rate of change of that curve on that interval. This line is called a secant line. The slope that we really want in Example 3 is the slope of the tangent line at x -3 that is the line that just "kisses" the graph at the point (-3,10) from undemeath the parabola. In general, the slope of the tangent line at a point gives us the slope of the curve at that point and is the instantaneous rate of change of the curve at that point. Interesting Fact Calculus was discovered independently in twa different parts of the world at roughly the same time. In 1674, Leibniz was in Germany trying to figure out how to find the slope of a tangent line to a curve at a single paint, while in 1666 in England, Isaac Newton was trying to determine the instantaneous rate of change of a particle in motion. It tums out that the solution to both of their questions was the Important Ideas Average Rate of Change Instantaneous Rate of Change slope of a Secant Line Slope of a Tangent Line Here's a hint of how our mathematical forefathers got around finding the slope ofthe tangent line with only one point. Example Given f(x) x +I, find the slope of the sec line between (a) x 3 and x 5 (b) 3 and x 2 (c) x 3 and x -0 (d) x --3 and -2 Example Given f(x). some small 4 approaches ze Example Using the sam want to find t
Explanation / Answer
We have given f(x)=x^2+1
a) x=-3 and x=5
at x=-3,y= f(-3)=(-3)^2+1=10
Let (x1,y1)=(-3,10)
at x=5,f(5)=26
Let (x2,y2)=(5,26)
we have two points (x1,y1)=(-3,10) ,(x2,y2)=(5,26)
Slope of the secant line between two points is
m=(y2-y1)/(x2-x1) =(26-10)/(5+3)=16/8 =2
m=2
b) when x=-3,f(-3)=10 and when x=2,f(2)=5
Let two points (x1,y1)=(-3,10) ,(x2,y2)=(2,5)
Slope of the secant line between two points is
m=(y2-y1)/(x2-x1)=(5-10)/(2+3)=-1
m=-1
c) when x=-3,f(-3)=10 and when x=0,f(0)=1
Let two points (x1,y1)=(-3,10) ,(x2,y2)=(0,1)
Slope of the secant line between two points is
m=(y2-y1)/(x2-x1)=(1-10)/(0+3)=-3
m=-3
d) when x=-3,f(-3)=10 and when x=-2,f(-2)=5
Let two points (x1,y1)=(-3,10) ,(x2,y2)=(-2,5)
Slope of the secant line between two points is
m=(y2-y1)/(x2-x1)=(5-10)/(-2+3)=-5
m=-5