Calculus AB - Slope of secant and tangent lines

Calculus AB ReviewSlopes 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:

Slope of a tangent line at one point:

Point-Slope Equation:

m=--------f(x₀)-f(x₁)x₀-x₁

m=--------f(h+x₀)-f(x₀)h

y-y₀=m(x-x₀)


1. f(x)=---

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

x-2

x-5


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

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

Our Line’s Equation

m=--------f(x₀)-f(x₁)x₀-x₁

Secant Line Slope Formula


1.a) Steps1.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:

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:

m=--------f(3)-f(4)3-4

f(3)=--- = -1/23-23-5

f(4)=--- = -24-24-5


1.a) Steps3.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:

4.So from here we’ll do basic simplification, which results in our answer:

m=--------(-1/2)-(-2)3-4

m = -1.5


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

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


Tangent Line Slope Formula

m=--------f(h+x₀)-f(x₀)h


1.b) Steps1.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:

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):

m=--------f(2+h)-f(2)h

f(2+h)= --- = h/(h-3)2+h-22+h-5 f(2)=--- = 02-2

2-5


1.b) Steps3.So, now we’ll look at the equation with the

function values plugged in (f(h+2)=h/(h-3), f(2)=0).

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.

m=--------h/(h-3)h = --------h

h(h-3) = --------1h-3

m = -- = -1/3 as h approaches 010-3

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.

h


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

-1/3

Tangent Line’s Slope

Point-Slope Formula

y-y₁ = m(x-x₁)

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



1.c) Steps1.To begin solving for the tangent line’s

equation, we substitute x=2 in to our line’s equation:

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):

f(x)= --- = 02-22-5

y-0 = -1/3(x-2)


1.c) Steps3.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) Steps1.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:

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:

m=--------f(-1)-f(2)3-4

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


2.a) Steps3.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:


m=--------(6)-(3)-1-2

m = -1


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


Tangent Line Slope Formula

m=--------f(h+x₀)-f(x₀)h

f(x)=2x²-3x+1


2.b) Steps1.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:

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):

m=--------f(2+h)-f(2)h

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



function values plugged in.

4.Now with our equation m=2h+5, 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.

m=--------2h²+3+5h-3h = --------2h²+5h

h = --------2h+51

m = 0+5 = 5 as h approaches 0

h


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

5

Tangent Line’s Slope

Point-Slope Formula

y-y₁ = m(x-x₁)


f(x)=2x²-3x+1


2.c) Steps1.To begin solving for the tangent line’s

equation, we substitute x=2 in to our line’s equation:

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)

y=2(2)²-3(2)+1 = 3


2.c) Steps3.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


Vavg=--------s(t₀)-s(t₁)

t₀-t₁

Velocity Average (Secant Line Slope) Formula

s(t)= 2t²-5t+1


3.a) Steps1. 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:

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:

Vavg=--------s(0)-s(10)0-10

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


3.a) Steps3.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:


Vavg=--------(1)-(151)0-10

Vavg = 15 feet per second


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


Instantaneous Velocity (Tangent Line Slope) Formula

Vinst= ---------s(h+t₀)-s(t₀)h

s(t)= 2t²-5t+1


3.b) Steps1. 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:

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):

Vinst=--------s(3+h)-s(3)h

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



function values plugged in.

4.Now with our equation Vinst=2h+7, 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 instantaneous velocity at 3 seconds.

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

h = --------2h+71

Vinst = 0+7 = 7 feet per second


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Calculus AB - Slope of secant and tangent lines