In Exercises 5–10, find an equation for the tangent line to th...

Question

In Exercises 5–10, find an equation for the tangent line to the curve at the given point. Then sketch the curve and tangent line together.

y = 4 − x², (−1, 3)

Accepted Answer

At the specified point, determine the equation of the tangent line to the curve. Also, plot the curve in the tangent line in the same coordinate system, where we have Y equals 11 minus X squared at the 0.32. We are also given a coordinate plane to plot our equations on. Now, to solve this, let's first find the equation of the tangent line. We will use point slope form, Y minus Y1 equals M multiplied by X minus X1. Now to find their slope M, we will use the difference quotient. To find the derivative of the function. Limit as H approaches 0 of F of X plus H minus F of X, all divided by H. Now our X value will be 3, which means we can find F of 3 plus H first. That's going to be 11 minus. 3 plus H, all squared. Now, let's expand this. We have 11 minus. B2 + 2 multiplied by 3, multiplied by H plus H2. That's our expansion for 3 + H2d. We have 11 minus in parentheses 9. Plus 6 H. Plus each squirt It's 11 9, minus 6 H minus H squad. Or 2 minus 6 H minus H squared. We also have F3. Which will be 11 Minus 3 squad, which is just 2. Now let's plug this into our limit definition formula. M equals the limit. As H approaches 0. Of 2 Minus 6 H minus H squared. Minus F of X, which is just 2, all divided by H. Our twos will cancel each other out, leaving us with a limit, as each approaches 0. Of -6 H minus H2 divided by H, which is the same as saying H multiplied by -6 minus H, all divided by H. Our Hs cancel as well. Leaving us with the limit As H approaches 0, Of -6 minus H, which is just 6 minus 0. Or just giving us -6. That will be the slope of our tangent line. Now let's plug this into our point slope form. Y minus Y1 equals M multiplied by X minus X1. We have Y minus 2 equals -6 multiplied by X minus 3. Giving us Y minus 2 equals -6 X plus 18. Adding 2 on both sides, gives us the tangent line, Y equals -6X plus 20. This is our tangent line equation. Now, let's graph this on our plane. We have Y equals 11 minus X2. Our Y intersects. We let X equals 0. This gives us, Y equals 11 minus 0 squared, or just 11, giving us a point of 0. 1 For the X innercents. We'll let Y equals 0, giving us 0 equals 11 minus X squared. Or -1 equals x2. Dividing by -1. Gives us X2 equals 11, or X equals plus or minus the square root of 11, when we take the square root of both sides. We have 2 X intercepts then, being in the square root of 11, 0. And square root 11, 0. This is also a downwards facing parabola, because the negative in front of the X squared. So we can now graph our functions. For F of X we have the intercept. 11 0. And square root of 11 0. Which is approximately between 3 and 4. And square root 11, 0. We will plot both of those. And make a smooth parabola between the points. Now let's find our tangent line. We're looking for the tangent line at the 0.32. It will actually adjust this parabola just a little bit to better fit. This function. You should have the line going through the 0.32. And so now, We can make a tangent line. We'll make this a different color. And a tangent line is 6 X + 20. Now, we already know we have a point at 32. Let's find a different point. Let's say X equals 0. We get Y equals -6 multiplied by 0, plus 20, which is just 20, giving us a point of 0 20. So, if we go back to our graph, we plot the point at 020, and then just draw a straight line between the two points. This gives us our function and this tangent line at the 0.32. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

In Exercises 5–10, find an equation for the tangent line to the curve at the given point. Then sketch the curve and tangent line together.

y = 4 − x², (−1, 3)

Key Concepts

Derivative

Point-Slope Form of a Line

Graphing Functions and Tangent Lines

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