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Determine the equations of the tangent lines from the point (0,13) to the curve $y=4-x^{2}$

$$y=-6 x+13, y=6 x+13$$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 2

Derivatives Rules 1

Derivatives

Missouri State University

Baylor University

University of Michigan - Ann Arbor

Idaho State University

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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03:37

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03:56

Find the slopes of the tan…

Okay, so here we are going to be using some basic calculus, mainly derivatives, to find the tangent lines that pass through the 0.0 13 and are tangent to the curve y equals four minus X squared. To start. Let's identify that our equation for our curve is F of X equal to four minus x squared. We want to find that derivative to give us the slope of our tangent lines. So F prime of X is equal to negative two X using some arbitrary point. Let's call it a. We know that we'll have a point of a f of A and a slope, which is equal to negative to a plugging a back into our ffx equation. Here we find that our generic coordinate point will be a four minus a squared. We're gonna be using both these coordinate points and our slope as we progress here. So now let's start by using a point slope form, which is recalled y minus y one is equal to m times X minus X one, plugging our values in of X and y that we just identified here. Let's do that. So we have y minus four minus a squared is equal to R M value of negative to a times X minus a simplifying we have while I minus four plus a squared is equal to negative to a X plus to a squared solving for y we get Y is equal to negative two x plus a squared plus four. We want to remember that going forward because we'll be using it now. We're gonna take our point of 0 13 that we're using we know are tangent lines either start or pass through there. Plug those values into our Y equation. We just found doing this. We get that. Why? Which is 13 is equal to negative to a our X value of zero plus a squared plus four. We're doing this so that we can find the value of a simplifying. We get that 13 is equal to a squared plus four finding in here a squared is equal to nine, making a equal to plus or minus three. With this a value, we're now going to plug that back into our Y equation again. In doing that we have why is equal to negative two times. Let's use positive three to start X was a squared, which is three plus four. Simplifying this, we get that Y is equal to negative six X plus 13. That gives us one of our tangent lines. Finding the other one. We could plug negative three into these equations. So suppose we did negative three. Instead of positive three and doing that, we would find our second tangent line, which is why equals six X Plus 13.

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