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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 $P$.$$f(x)=x^{2}+3 \quad P(1,4)$$

A. (1,4)C. 2

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 1

Slope of a Curve

Derivatives

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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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all right here. We're gonna be using some basic calculus to graph the function and determine a determinant slope at a particular point. To do that we'll be using are known slope formula where M is equal to y tu minus y one all over x two minus x one. Start by sketching a graph of this function. We know that's going to have a minimum at y equals three and sketching that it looks something like this. That's what an X squared function typically looks like were given the 0.1 comma four, which would sit right in here. And we want to determine what the slope is at that point. To do that, what we would do is draw a tangent line through it and what we need to find another point somewhere on that tangent line. To do that, we're going to start plugging some values into our function that was given to us, and we want to keep this point as close as possible to our 0.0.1 comment forward To ensure the accuracy of our finding, let's go ahead and use a value of X is equal to, let's say, 1.1 Let's plug that into our function. Plus three. If we calculate that out, we'll end up finding that will get a Y value equal to 4.2 Now that we have both the point that was given to us and the point that we just found we can start to plug these values into our slope formula up here, Let's do that and say M is equal to R Y to 4.0 to minus Y one of four, all over x two, which we found was 1.1 minus one. In doing this, we end up getting to for that. This tells us that our slope at point 14 on this curve is equal to two.

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