Calculus AB - Slope of secant and tangent lines

Calculus AB Review
Slopes of non-linear secant/tangent
lines, and point-slope formula

Kenyon Hundley

Tuesday, September 25, 12

Important Formulas

Slope of a secant line at two points:
f(x₀)-f(x₁)
m=--------
x₀-x₁
Slope of a tangent line at one point:
f(h+x₀)-f(x₀)
m=--------
h

Point-Slope Equation:
y-y₀=m(x-x₀)


x-2
1. f(x)=---
x-5

a) find the slope of the secant line
between x = 3 and x = 4

b) find the slope of the tangent line
at x = 2

c) find the equation of the tangent
line at x = 2


1.a) find the slope of the secant
line between x = 3 and x = 4

Our Line’s Equation
x-2
f(x)=---
x-5

Secant Line Slope Formula

f(x₀)-f(x₁)
m=--------
x₀-x₁


1.a) Steps
1. Using the secant line slope formula we can
plug in the two values of ‘x’ that are given
to us. We’ll just say x₀=3 and x₁=4. This looks
like: f(3)-f(4)
m=--------
3-4
2. So on the top we see we have f(3) and f(4),
which we interpret as “plug in 3 and 4 as x-
values into the function and the solve for
the y-value (or f(x) in other words)”. When
we plug 3 and 4 in as x-values to our given
equation, it looks like this:
3-2 4-2
f(3)=--- = -1/2 f(4)=--- = -2
3-5 4-5

1.a) Steps
3. So now that we know f(3)=-1/2 and f(4)=-2,
we can plug these values back into our
secant slope equation which will look like:
(-1/2)-(-2)
m=--------
3-4
4. So from here we’ll do basic simplification,
which results in our answer:

m = -1.5


1.b) find the slope of the
tangent line at x = 2

x-2
f(x)=---
x-5

Tangent Line Slope Formula

f(h+x₀)-f(x₀)
m=--------
h


1.b) Steps
1. Using the tangent line slope formula we’ll
plug in the value of ‘x’ that is given to us.
Since x=2, this looks like:
f(2+h)-f(2)
m=--------
h
2. So in order to find the slope, we want to
simplify the equation to a point where, if we
plugged 0 in for ‘h’, we would not result in
an undefined equation (0 as the denominator).
We’ll start by solving the functions that are
defined in the numerator, f(2+h) and f(2):

2+h-2 2-2
f(2+h)= --- = h/(h-3) f(2)=--- = 0
2+h-5 2-5


1.b) Steps
3. So, now we’ll look at the equation with the
function values plugged in (f(h+2)=h/(h-3),
f(2)=0).
h/(h-3) h 1
m=-------- =
h
-------- = --------
h(h-3) h-3

4. Now with our equation m=1/(h-3), we can see
that plugging in 0 for ‘h’ won’t result in an
undefined value so we can use this to solve
for the slope. h
1
m = -- = -1/3 as h approaches 0
0-3
We write the answer like “m = -1/3 as h approaches 0” because we cannot find
the exact slope of the tangent line at a given point, but we can get really
close to it by plugging in 0 for ‘h’, or a decimal value extremely close to 0.


1.c) find the equation of the

x-2
f(x)=---
x-5

Point-Slope Formula

y-y₁ = m(x-x₁)

Tangent Line’s Slope

-1/3


1.c) Steps
1. To begin solving for the tangent line’s
equation, we substitute x=2 in to our line’s
equation: 2-2
f(x)= --- = 0
2-5

2. So now we know that 0 is x’s corresponding y-
value. We’ll use (2,0) as our x₁ & y₁ in the
point-slope equation, and also -1/3 as the
slope (we found the slope of the same
tangent line in 1.b):
y-0 = -1/3(x-2)


1.c) Steps
3. Now we basically just simplify the equation
to slope-intercept form, then we have our
answer:
y = -1/3x+2/3


2. f(x)=2x²-3x+1

a) find the slope of the secant line
between x = -1 and x = 2

b) find the slope of the tangent line
at x = 2

c) find the equation of the tangent
line at x = 2


2.a) Steps
1. Using the secant line slope formula we can
plug in the two values of ‘x’ that are given
to us. We’ll just say x₀=-1 and x₁=2. This looks
like: f(-1)-f(2)
m=--------
3-4
2. So on the top we see we have the functions
f(-1) and f(2). When we plug -1 and 2 in as
x-values to our given equation, it looks like
this:

f(-1)=2(-1)²-3(-1)+1= 6
f(2)=2(2)²-3(2)+1= 3

2.a) Steps
3. So now that we know f(-1)=6 and f(2)=3, we
can plug these values back into our secant
slope equation which will look like:
(6)-(3)
m=--------
-1-2

m = -1


2.b) find the slope of the


f(x)=2x²-3x+1

Tangent Line Slope Formula

f(h+x₀)-f(x₀)
m=--------
h


2.b) Steps
1. Using the tangent line slope formula we’ll
plug in the value of ‘x’ that is given to us.
Since x=2, this looks like:
f(2+h)-f(2)
m=--------
h
2. So in order to find the slope, we want to
simplify the equation to a point where, if we
plugged 0 in for ‘h’, we would not result in
an undefined equation (0 as the denominator).
We’ll start by solving the functions that are
defined in the numerator, f(2+h) and f(2):

f(2+h)=2(2+h)²-3(2+h)+1=2h²+3+5h
f(2)=2(2)²-3(2)+1=3

2.b) Steps
function values plugged in.

2h²+3+5h-3 2h²+5h 2h+5
m=-------- =
h
-------- = --------
h 1
4. Now with our equation m=2h+5, we can see that
plugging in 0 for ‘h’ won’t result in an
for the slope.
h

m = 0+5 = 5 as h approaches 0


2.c) find the equation of the


f(x)=2x²-3x+1

Point-Slope Formula

y-y₁ = m(x-x₁)

Tangent Line’s Slope

5


2.c) Steps
1. To begin solving for the tangent line’s
equation, we substitute x=2 in to our line’s
equation:
y=2(2)²-3(2)+1 = 3

2. So now we know that 3 is x’s corresponding y-
value. We’ll use (2,3) as our x₁ & y₁ in the
point-slope equation, and also 5 as the
slope (we found the slope of the same
tangent line in 2.b):
y-3 = 5(x-2)


2.c) Steps
3. Now we basically just simplify the equation
to slope-intercept form, then we have our
answer:
y = 5x-7


3. If an object travels a
distance of s = 2t²-5t+1,
where s is in feet and t is in
seconds,find:

a) the average velocity of object
within the first 10 seconds.

b) the instantaneous velocity of the
object at t = 3 seconds.


3.a) the average velocity of
object within the first 10 feet


s(t)= 2t²-5t+1
Velocity Average (Secant Line
Slope) Formula
s(t₀)-s(t₁)
Vavg=--------
t₀-t₁


3.a) Steps
1. In this problem, we basically use the same rules and
formula as the secant line slope formula, except we
use different notation. We can plug 10(seconds) and
0(seconds) for t₀ and t₁ in order to find our average
velocity during the first ten seconds. We’ll say t₀=0
and t₁ =10. This looks like:
s(0)-s(10)
Vavg=--------
0-10
2. So on the top we, once again, see that we have
functions; s(0) and s(10) are our functions. After
plugging them into our line’s equation, we get this:

s(0)= 2(0)²-5(0)+1 = 1
s(10)= 2(10)²-5(10)+1= 151

3.a) Steps
3. So now that we know s(0)=1 and f(10)=151,
we can plug these values back into our
average velocity equation which will look
like:
(1)-(151)
Vavg=--------
0-10


Vavg = 15 feet per second


3.b) the instantaneous velocity
of the object at t = 3 seconds


s(t)= 2t²-5t+1
Instantaneous Velocity (Tangent
Line Slope) Formula
s(h+t₀)-s(t₀)
Vinst= ---------
h


3.b) Steps
1. To discover the instantaneous velocity at t=3 seconds, we
basically use the tangent line slope formula but with
different notation. We plug in the given value of 3 in
for “t” in the Vinst equation. This looks like:
s(3+h)-s(3)
Vinst=--------
h
2. So, (like the tangent line slope equations) in order to
find the instantaneous velocity, we want to simplify the
equation to a point where, if we plugged 0 in for ‘h’, we
would not result in an undefined equation (0 as the
denominator). To start we’ll solve the functions that are
defined in the numerator, s(3+h) and s(3):

s(h+3)= 2(h+3)²-5(h+3)+1= 2h²+7h+4
s(3)= 2(3)²-5(3)+1= 4

3.b) Steps
function values plugged in.

2h²+7h+4-4 2h²+7h 2h+7
Vinst=-------- =
h
-------- = --------
h 1

4. Now with our equation Vinst=2h+7, we can see
that plugging in 0 for ‘h’ won’t result in an
for the instantaneous velocity at 3 seconds.
Vinst = 0+7 = 7 feet per second


Calculus AB - Slope of secant and tangent lines

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