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Help with tangent lines?

Topic: Help with tangent lines?
June 20, 2019 / By Claretta
Question: I'm going to be taking Calculus next year, and I got a summer packet that reviews a few basic concepts we'll need. It mentions something about calculating the slope of a tangent line, and my memory's kind of fuzzy on that. It says to consider the graph of y = x^2, and the point (2,4) and find the slope of the tangent line (so I need help on calculating tangent lines in general) It also says "Find the slope of the line joining (2,4) and (2 + h , f(2 + h)) in terms of the nonzero number h". How would you go about doing that? (Is the f(2 + h) a typo?) It also has 2 questions about evaluating functions at a given value that I haven't figured out. It says f(x) = x^3 and then says the following: f(x + delta x) - f(x) and all of it is divided by delta x The delta x is what's throwing me off. Then another question says f(x) - 3x - 1 and then says f(x) - f(1) all over x - 1. I think you would put 3x - 1 - 2 / 1, which is 3x - 3, correct? Thanks in advance, everybody.

Best Answers: Help with tangent lines?

Basmath | 10 days ago
Ok so in this case, delta x and h mean the same thing: the change in the x value. for your first problem, you have to use the formula f((x+h) - f(x))/h as h approaches 0. so for this problem, you would do (x+h)^2 (substituting x+h in the place of x) - x^2 / h (x2 + 2xh + h2 - x2) / h 2xh + h2 / h 2x + h as h approaches 0, the equation becomes 2x. this equation will give the slope of the tangent line if you plug in an x value for it. so if you plug in the 2 in (2, 4) you get 4. you can then use point slope form to get the entire tangent line equation: y - 4 = 2(x - 2) For your second problem, the delta x functions the same way as h, it shows the change in x from its orginal position. so you can do the same thing as the first problem. if it doesn't give you points, you can just stop after you take the limit of the equation. I'm not sure what you mean by the third equation. I think you tried to type f(x) = 3x - 1. If that's what you meant to type, the question is asking you to do the same thing as questions 1 and 2, but with a different formula: f(x) - f(h) / x- h, so you solved it correctly. The answer you get is the same as if you had done the x + h method with point slope, but a bit faster. Hope that helps.
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Start with the y = x³: y = x³ Take the first derivative: y ' = 3x² Looking back at the y = 3x + k, you can see that the slope of the tangent line is 3. Remember that the first derivative is the slope of the tangent line. So you can set 3=3x²: 3 = 3x² 1 = x² x = ±1 So either x = 1 or x = -1 will work as a point for x. Going back to the beginning when x = 1, y = 1. So the point we'll use to solve for k is (1,1): 1 = 3(1) + k 1 = 3 + k k = -2 <==Answer And there's actually a 2nd possibility. If you were wondering what would have happened if you chose -1 as the point. Well, when x= -1, y = -1. So: -1 = (-1)(3) + k -1 = -3 + k k = 2 <==Other answer
i can try to help you....i hope you have answers :P y=x^3 and y=3x+k let them equal each other as you find the tangent when the lines intersect, and solve for k.