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=√7−x, P(−2,3)

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 square root of 78 minus 7 X and point S is 2.8. Essentially, we're going to identify the equation of the tangent line in a form of Y equals MX + B. So we want to find the slope and the Y intercept. First of all, let's identify the general equation of the slope using the limit definition. We're going to identify the limit as H approaches 0 of F at X plus H minus F of X divided by H. Here F is our function, so all that we have to do is evaluate F at X + H, which is square root of 7C8 minus 7 multiplied by X plus H. And then our F of X is given, right? We know that it is equal to square root of 78 minus 7 X. So let's substitute those values into the formula. This gives us limit as H approaches 0 of square root. Of 78 minus 7 multiplied by X plus H. We are subtracting square root of 78 minus 7 X, and all of that is divided by H. Now how do we solve for this limit? Well, essentially we need to multiply and divide both sides of the fraction by the conjugate of the numerator, otherwise we simply end up with an indeterminate form of 0 divided by 0. So we're going to continue and we're going to say that the slope. is equal to the limit as H approaches 0, and now let's remember what a conjugate is. We're going to write the original numerator, square root of 78 minus 7 multiplied by X plus H. Minus square root of 78 minus 7 X. And now the conjugate is basically the same term with a positive sign instead of the negative sign between the two radicals. So we have to take square root off. 78 minus 7 multiplied by X plus H. Less square root off. 78 minus 7 X. So that's how we're conjugate, and because we're multiplying the numerator by that conjugate, we also have to multiply the denominator. So we get H multiplied by. Square root of 78. -7 multiplied by X plus H. Plus square root of. 78 minus 7 X. And now we can multiply. This gives us limit. As H approaches 0, for the numerator, we can see that we have a factored form of the difference of squares, so we can just square the first term. We get 78 minus 7 multiplied by X plus H, and then we're subtracting the square of the second term. So we're subtracting 78 minus 7 X. All of that is divided by the denominator. We are now going to combine the like terms and the numerator. We're also going to apply the distributive property. So we get 78 minus 78, those terms will cancel each other out, then we get -7 X minus 7 H. And then we are subtracting negative 7X. Basically this means that we're adding 7 X, and we can now see that -7X plus 7X, this gives us 0, so we only end up with negative 7H. And for our denominator, nothing changes, we still have the same product between the age term and the conjugate, right? So essentially we can see that our age can be canceled out and then we're going to adjust the denominator. So we only have the conjugate remaining rights. We can essentially eliminate the whole, so we can get rid of age, and then we can just write the conjugate in the denominator. Now from here, we're going to substitute H. Equals 0. Let's also add the conjugate back to the previous form. And we're going to identify the equation of the slope. So we get -7 divided by. Now, if H is 0, we basically get two exactly same radicals, so we can say in the denominator we have two square root of. 78 minus 7 X. That's the general equation of the slope, and now, because it's dependent on X, we can identify the slope at.2 by substituting 2 into the equation, which gives us -7. Divided by 2 square roots of. 78 minus 7 multiplied by 2. This gives us -7 divided by 2 multiplied by 8, right? So -7 divided by 16. Well done, so now that we have our slope, our next goal is to identify the intercept. What we can do is basically get rid of all of this work. And continue with the equation. We know that our slope is equal to -7 divided by 16, and now we can use the point slope formula. Let's recall that Y minus Y1 is equal to M multiplied by X minus X1. In this context, our X1 is 2, that's given within our point, and Y1 is equal to 8. So we get Y minus 8 is equal to -7 divided by 16 multiplied by X minus 2. So Y equals -7 divided by 16 X, we're applying the distributive property. Plus 7 multiplied by 2, this gives us 14, so we get 14 divided by 16, and then we're adding 8 to both sides. So now let's simplify. Y equals -7 divided by 16 X. Less. 7 divided by 8, we can simplify the fraction, dividing both sides by 2, and then 8 is just 64 divided by 8. Combining the like terms, we end up with Y equals -7 divided by 16 multiplied by X plus 71 divided by 8, which is our final equation for this problem. 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=√7−x, P(−2,3)

Key Concepts

Derivative

Tangent Line

Point-Slope Form

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