Question 1208077: The following question is from my college algebra textbook. It is preview of calculus 1.
Let m_sec = slope of the secant line.
I am to use m_sec = [(f(x + h) - f(x)]/h. It is called the difference quotient of f.
Given f(x) = 2x^2 + x:
A. Express the slope of the secant line of the given function in terms of x and h. Be sure to simplify your answer.
I found m_sec to be 4x + 2h + 1.
B. Find m_sec for h = 0.5 at x = 1. What value does m_sec approach as h tends to 0?
I don't understand part B.
C. Find the equation of the secant line at x = 1 with h = 0.01.
I don't know how to find the equation of the secant line for part C.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part A is correct. Nice work.
--------------------------------------------------------------------------
Part B
m_sec = 4x+2h+1
m_sec = 4(1)+2(0.5)+1
m_sec = 6
As h gets closer to 0, the 2h will also get closer to 0.
This will mean the 2h is so small that it's practically zero (even if we technically don't land on this value exactly), and it goes away.
Therefore m_sec approaches 4x+1 as h tends to 0.
A bit of a spoiler alert: You'll learn later in Calculus that this is the derivative of 2x^2+x.
The derivative is very useful in many applications. One of which is finding the slope of a tangent line.
--------------------------------------------------------------------------
Part C
m_sec = 4x+2h+1
m_sec = 4(1)+2(0.01)+1
m_sec = 5.02 is the slope of the secant line at x = 1 where h = 0.01
f(x) = 2x^2 + x
f(1) = 2*1^2 + 1
f(1) = 3
The point (x,y) = (1,3) is on the f(x) curve.
Meaning that x = 1 and y = 3 pair up together.
y = mx+b
3 = 5.02*1+b
3 = 5.02+b
b = 3-5.02
b = -2.02
The equation of the secant line is y = 5.02x-2.02
To help verify you can plug x = 1 into this equation and you should get y = 3.
Another approach to finding the secant line is to use the point-slope template.
y-y1 = m(x-x1)
y-3 = 5.02(x-1)
y-3 = 5.02x-5.02
y = 5.02x-5.02+3
y = 5.02x-2.02
|
|
|
