Question 1208077
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Part A is correct. Nice work.


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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 <font color=red>4x+1</font> 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.


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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 <font color=red>y = 5.02x-2.02</font>
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
<font color=red>y = 5.02x-2.02</font>
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