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Topic: Help with tangent lines?**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.

June 20, 2019 / By Claretta

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.

👍 228 | 👎 10

Did you like the answer? 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.

there is no typo, h or delta is some infinitely small positive value. you already have everything presented; f(x)=x^2 then f(x+h)=(x+h)^2=x^2+2xh+h^2 because h is REALLY small (i guess you don't know yet about limits), when you square it, you get even SMALLER number. so small in fact that h^2 is neglected. then you find slope from [f(x+h)-f(x)]/[x - (x+h)] on the top, x^2 cancels, on the bottom x cancels and you get just 2xh/h which is 2x if you know derivatives you would know how to get this directly. what i derived here is work done "by definition" (you do this once or twice and move on and just use derivatives)

👍 90 | 👎 3

i will get the tangent as i don't comprehend what do u mean by using widely used line y=mx+b to get m get f' at this element f(x)=(x^2+3)/(2x) = x/2 + 3/2x f'(x)=a million - 3/(2x^2) f'(a million)=a million-3/2= - a million/2 so m= -a million/2 y= -a million/2 x +b to get b use the element (a million,2) a million = (-a million/2 )*2 +b a million=-a million+b b=2 so the line equation of the tangent is y= -x/2 +2

👍 81 | 👎 -4

y' = lim(h->0) (f(x+h) - f(x)) / h y' = lim(h->0) (1/(x+h-1) - 1/(x-1)) / h At this point multiplying by the LCD to get rid of the complex fraction will simplify this a lot. y' = lim(h->0) ((x-1) - (x+h-1)) / h(x-1)(x+h-1) y' = lim(h->0) (-h) / h(x-1)(x+h-1) y' = lim(h->0) -1 / (x-1)(x+h-1) y' = -1 / (x-1)(x-1) y' = -1 / (x-1)^2 Since the slope of the tangent line must be -1: -1 = -1 / (x-1)^2 (x-1)^2 = 1 x-1 = ±1 x = 1 ± 1 x = 0, 2 The two tangent lines touch the function at x=0 and x=2. x = 0 y = 1/(x-1) = 1/(-1) = -1 (0,-1) x = 2 y = 1/(x-1) = 1/1 = 1 (2,1) Each tangent line is of the form y=mx+b, and m=-1 as given by the problem. y = -x + b -1 = -(0) + b b = -1 y = -x - 1 y = -x + b 1 = -(2) + b 3 = b y = -x + 3 So the two tangent lines are y=-x-1 and y=-x+3.

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