For the function ƒ(x) = x2 find the slope of secants from the point (2, 4) to each of the following points.
(3, __ )
(2.5 , __ )
(2.1 , __ )
(2.01, __ )
(2.001, __ )
Using the pattern from question 1, what do you think the slope of a tangent at the point (2, 4) would be? Explain you answer.
I will do the third point for you
x = 2.1 so f(2.1) = (2.1)^2 = 4.41
so slope of line between (2,4) and (2.1,4.41) is (4.41-4)/(2.1-2)
= 4.1
Now you do the rest in the same way.
Do you notice what is happening?
To find the slope of the secants from the point (2, 4) to each of the given points, we need to calculate the slope using the formula:
Slope = (ƒ(x2) - ƒ(x1)) / (x2 - x1)
Let's calculate the slope for each point:
(3, __):
Slope = (ƒ(3) - ƒ(2)) / (3 - 2)
= (9 - 4) / 1
= 5
(2.5, __ ):
Slope = (ƒ(2.5) - ƒ(2)) / (2.5 - 2)
= (6.25 - 4) / 0.5
= 2.25 / 0.5
= 4.5
(2.1, __ ):
Slope = (ƒ(2.1) - ƒ(2)) / (2.1 - 2)
= (4.41 - 4) / 0.1
= 0.41 / 0.1
= 4.1
(2.01, __ ):
Slope = (ƒ(2.01) - ƒ(2)) / (2.01 - 2)
= (4.0401 - 4) / 0.01
= 0.0401 / 0.01
= 4.01
(2.001, __ ):
Slope = (ƒ(2.001) - ƒ(2)) / (2.001 - 2)
≈ (4.00040001 - 4) / 0.001
≈ 0.00040001 / 0.001
≈ 0.40001
Using the pattern from the calculations, we can observe that as the x-values approach 2 (the given point), the slopes of the secants appear to approach 4. Therefore, we can infer that the slope of the tangent line at the point (2, 4) would be approximately 4. This can be explained by the fact that the derivative of the function ƒ(x) = x^2 is 2x, and when x is equal to 2, 2x is equal to 4. Hence, the slope of the tangent line at (2, 4) is 4.
To find the slope of secants from the point (2, 4) to each of the given points, we need to calculate the slope using the formula:
Slope = (change in y) / (change in x)
Let's calculate the slope for each point:
1. (3, __):
Slope = (y2 - y1) / (x2 - x1)
= (__ - 4) / (3 - 2)
Since we don't have the value for y2, we cannot calculate the slope.
2. (2.5, __):
Slope = (y2 - y1) / (x2 - x1)
= (__ - 4) / (2.5 - 2)
Again, since we don't have the value for y2, we cannot calculate the slope.
3. (2.1, __):
Slope = (y2 - y1) / (x2 - x1)
= (__ - 4) / (2.1 - 2)
Still, we don't have the value for y2, so we cannot calculate the slope.
4. (2.01, __):
Slope = (y2 - y1) / (x2 - x1)
= (__ - 4) / (2.01 - 2)
Similarly, we don't have the value for y2, so we cannot calculate the slope.
5. (2.001, __):
Slope = (y2 - y1) / (x2 - x1)
= (__ - 4) / (2.001 - 2)
Once again, we don't have the value for y2, so we cannot calculate the slope.
Now, let's discuss the pattern we see. As the x-coordinate of the second point gets closer and closer to 2, but never actually reaching it, the slope of the secant lines seems to be approaching a certain value.
In this case, the slope of the tangent line at the point (2, 4) can be approximated by calculating the slope of secants whose x-coordinate of the second point is getting closer and closer to 2.
We can express this pattern using calculus by taking the derivative of the function f(x) = x^2 and evaluating it at x = 2. The derivative of f(x) = x^2 is f'(x) = 2x. Evaluating f'(x) at x = 2 gives us the slope of the tangent line at x = 2.
f'(2) = 2 * 2 = 4
Thus, the slope of the tangent line at the point (2, 4) is 4.