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For the function​f(x)=-3x^2make a table of slopes of secant lines and make a conjecture about the slope of the tangent line at x=2.Interval [1,2]

Question

For the function

​f(x)=-3x^2

make a table of slopes of secant lines and make a conjecture about the slope of the tangent line at x=2.

Interval [1,2]

check_circleAnswer
Step 1

It is known that the slope of a secant line between points (a, f(a)) and (b, f(b)) is given by:

f(b)-f(a)
b-a
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f(b)-f(a) b-a

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Step 2

Now, evaluate slopes of secant line at different intervals between the provided interval [1,2]:

at 1,2
-2(9)
at 1.5,2
f(2)-(L.5)-12-(-6.75)
f(2)-f(1)12-(-3
2-1
1
-5.25
-10.5
2-1.5
0.5
0.5
at [1.9,2
f(2)-f(1.9) 12-(-10.83)
-1.17_11.7
2-1.9
0.1
0.1
at 199,2]
f (2)-f(1.99)-12-(-11.8803) -0.1197
--11.97
2-1.99
0.01
0.01
[1.999,2
at
f(2)-f(1.999)-12-(-
2-1.999
-11.988003)-0.011997
=-11.997
0.001
0.001
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at 1,2 -2(9) at 1.5,2 f(2)-(L.5)-12-(-6.75) f(2)-f(1)12-(-3 2-1 1 -5.25 -10.5 2-1.5 0.5 0.5 at [1.9,2 f(2)-f(1.9) 12-(-10.83) -1.17_11.7 2-1.9 0.1 0.1 at 199,2] f (2)-f(1.99)-12-(-11.8803) -0.1197 --11.97 2-1.99 0.01 0.01 [1.999,2 at f(2)-f(1.999)-12-(- 2-1.999 -11.988003)-0.011997 =-11.997 0.001 0.001

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Step 3

So, the table of slopes of seca...

Interval
[1.9,2]
-11.7
[1.999,2]
-11.997
[1,2]
-9
[1.5,21
-10.5
[1.99,2]
-11.97
Slope of secant line
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Interval [1.9,2] -11.7 [1.999,2] -11.997 [1,2] -9 [1.5,21 -10.5 [1.99,2] -11.97 Slope of secant line

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Tagged in

Math

Calculus

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