Estimate the instantaneous rate of change of h (t) = 2t² + 2 at the point t = −1.
in other words, choose x-values that are getting closer and closer to −1 and compute the slope of the secant lines at each value. then, use the trend/pattern you see to estimate the slope of the tangent line.

The instantaneous rate of change of h(t) = 2t² + 2 at the point t = −1 is -4.

Step-by-step explanation:

If two distinct points and lie on the curve , the slope of the secant line connecting the two points is

If we let the point approach , then Q will approach P along the graph f(x). The slope of the secant line through points P and Q will gradually approach the slope of the tangent line through P as

And this is the instantaneous rate of change of the function f(x) at the point .

From the information given, we know that the point P lies on the curve .

If Q is the point we can find the slope of the secant line PQ for the following values of t. Because we choose values that are getting closer and closer to −1.

Let the point P be and the point Q be . So,

Next, substitute the value of x in the formula of the slope

Do this for the other values of x.

Below, there is a table that shows the values of the slope.

From the table, as t approaches -1 from the left side (-0.9 to -0.9999), the slopes are approaching to -4 and as t approaches -1 from the right side (-1.1 to -1.0001), the slopes are approaching to -4. The value of the slope at P(-1,4) is then m = -4.

step-by-step explanation: