Slope of a Curve at a Point

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Question

Slope of a Curve at a Point

In Exercises 7–18, use the method in Example 3 to find (a) the slope of the curve at the given point P, and (b) an equation of the tangent line at P.

y=x³−3x²+4, P(2,0)

Accepted Answer

Welcome back, everyone. Calculate the slope of the curve at the specified point S and find the equation of the tangent line at S. Y equals X cubed minus 5 X2 + 18, and S is a point with X equals 3. Y equals 0. Let's begin with the slope, and for this problem we're going to use the definition of the slope. Using the limit definitions, we want to evaluate the limit as H approaches 0 of the function at X plus h. Then we are subtracting the function F of X and dividing by H. So that's the definition of the slope of the tangent line at any X. Let's evaluate each term limit as H approaches 0. 1st of all, we're taking F at X + H, meaning every X becomes X + H, right? So we get X + HQ. Minus 5 multiplied by X plus H2 + 18. And then we're going to subtract the original function in the numerator rights we're subtracting X cubed minus 5 X2 + 18. Let's add pri to indicate that we're subtracting the whole function. And then we're dividing by H. So from here we simply need to keep the first term. Let's begin with this step. We get limit as H approaches 0. Applying the formula for the cube of a sum of two terms, we're going to cube the first term, which gives us X cubed. We're going to add 3 multiplied by the square of X multiplied by H, so 3 X 2 H plus 3 multiplied by X multiplied by H2. Plus the cube of H. Then we are subtracting 5 multiplied by X plus H2. So here we're squaring x. Adding 2 multiplied by X multiplied by H and adding H2d. And then we can distribute the negative sign. We are subtracting X cube. Adding 5 x squad and subtracting 18, so flipping all of those signs. All of that is divided by H. So now let's start combining the like terms. We get limit. As H approaches 0. First of all, we can notice that we can cancel out X cubed minus X cubed. And then we will essentially need to distribute. Additionally, let's notice that we are missing + 18, so let's incorporate it into our equation to make sure that we are taking all of the terms that we have in the original limit. So let's notice that after we're taking -5. Multiplied by X + H2, we have that plus 18, right? So let's add plus 18. And we can see that also we can cancel out 18 and -18. So from now on, we can focus on distributing the remaining terms. We get limit as H approaches 0 of 3 X 2 H plus 3 XH2 + HQ. Minus 5 X2 minus 10 XH minus 5 H2 + 5 X2. All of that divided by H. Now, what can we see? Well, we have minus 5 X2 + 5 X2? Do we have any other like terms? Looks like we don't have any other like terms, so we're just going to perform factorization, right, because each term hash. So we get limit as H approaches 0 of H and 3x2 + 3 X H plus H2 minus 10 X. Minus 5 H. And that's it, right? For the numerator, and then we're dividing by H. We can clearly see that now we can cancel out the H term and evaluate the limit. Notice that H approaches 0. This means that every term containing H is going to be equal to 0, right, because we're multiplying by H in one form or another. So 5 H is going to approach 0, H squared is going to approach 03 XH is going to approach 0, so we simply end up with 3 X2 minus 10 X. 3 X2 minus 10 X. This is the general expression of the slope, right? 3 X2 minus 10 X. So let's Clear the screen and use this expression for the evaluation of the slope. We know that any slope M. Is equal to 3 x 2 minus 10 X. And the point of interest has X equals 3, right, because S contains coordinates 3.0. So the slope in this context at S is going to be equal to 3 multiplied by 3 squad minus 10 multiplied by 3. We're simply substituting X equals 3. Evaluating the result, we get 27 minus 30, and that's -3. So we can conclude that our first answer is slope equals. -3. We can label it as the first answer to this problem, and now we can focus on identifying the equation of the tangent line using the point slope formula, right? Let's recall that Y minus Y1 is equal to m multiplied by x minus X1. In this case, we can just rewrite it as Y minus Y S equals M multiplied by x minus Xs because we are given a points. So we get Y minus 0 because the Y coordinate of S is 0. is equal to the slope M, which is -3, multiplied by X minus 3, because the X coordinate of S is 3, right? This gives us Y equals -3 X plus 9. We're simply dis distributing, and that will be the equation of the tangent line. That's it. We have our answers and thank you for watching.

Slope of a Curve at a Point

In Exercises 7–18, use the method in Example 3 to find (a) the slope of the curve at the given point P, and (b) an equation of the tangent line at P.

y=x³−3x²+4, P(2,0)

Key Concepts

Derivative

Tangent Line

Point-Slope Form

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