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a) sketch the graph of the given function, and then draw the tangent line at the point $P$. (b) Using your sketch, approximate the slope of the curve at $P,$ (c) Use (1) to determine the exact value of the slope at $P$.$$f(x)=-2 x^{2}+3 x+3$$$\quad$$$P(2,1)$$

A. (2,1)C. -5

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 1

Slope of a Curve

Derivatives

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Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

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a) sketch the graph of the…

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Use the method in Example …

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all right here. We're gonna be using some basic calculus to determine the slope of the curve. At a given point, we are to sketch the function negative two X squared plus three x plus three. Doing this, we get a function that looks something like this. Here we go and were given the point to one. And we want to know what the slope is going through. That in order to do that, let's find a point very, very close to there so that we can get a second point on what would be a tangent line. Let's use the value X is equal to 2.1001 gets us pretty close, and let's now plug that into our function here. In doing that, we'll end up getting a Y value equal to 0.999499 Now that we have both this point and the point to one that was given to us, let's go ahead and plug these values into our known slope formula, where M is equal to y tu minus y one Oliver x two minus x one because now we're just calculating the slope of a straight line, which is our tangent line. Plugging these values in we've got em is equal to 0.999499 minus one all over 2.1 minus two. Calculating that we get em is equal to negative 5.1 So we conclude that we have a slope at 0.21 of the curve Negative two X squared plus three X Plus three is equal to about negative five. It's just a rough estimate, not quite rough, but it's an estimate because we chose a point as close as possible to to common one. But we could have gotten even closer if we wanted. We could have done 2.0 You know, we could keep going and get it as close as possible. We will continue to get a more accurate estimate, but it looks as though it's approaching negative five, so we can assume that we have a slope of negative five

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