This structured practice takes you through three examples of finding the equation of the line tangent to a curve at a specific point.
Log in BB8FN2187 6 years agoPosted 6 years ago. Direct link to BB8FN2187's post “Why do we need to know th...” Why do we need to know this? Couldn't we just take the derivative using derivative rules and be faster/easier? • (20 votes) ( ͡° ͜ʖ ͡°) 6 years agoPosted 6 years ago. Direct link to ( ͡° ͜ʖ ͡°)'s post “Yes, derivative rules are...” Yes, derivative rules are faster and easier than using the "first principles" definition of the derivative (with limits and the definition of slope), but the only reason we have those rules is because of that definition of the derivative. In practice, you do use the derivative rules, not the first principles definition, to find derivatives, but it's really important to understand the intuition of why we can even find instantaneous slope in the first place (which was a nonsensical idea to people more recently than you would think). (141 votes) Abel Romo 5 years agoPosted 5 years ago. Direct link to Abel Romo's post “y = -10(x + 5) + 28y = -...” y = -10(x + 5) + 28 • (37 votes) djschrock00 6 years agoPosted 6 years ago. Direct link to djschrock00's post “I found using point-slope...” I found using point-slope form to solve for the final equation to be easier than this. I entered my answer to the third question as -10x-22 and that is correct (if you simplify the answer given, it is the same) but it might be confusing to others. • (29 votes) Ivan Korniienko 3 years agoPosted 3 years ago. Direct link to Ivan Korniienko's post “Deat Khan Academy, please...” Deat Khan Academy, please communicate it clearer that the "Example 3: Finding the equation of the line tangent" MUST contain "y=" at the front of the answer. I nearly lost my mind trying to understand where was my arithmetic wrong. • (21 votes) henrydavidthomas 3 years agoPosted 3 years ago. Direct link to henrydavidthomas's post “it asks for the equation ...” it asks for the equation of a line. a line equation is typically of the form y=mx+b (17 votes) Mihrr 6 years agoPosted 6 years ago. Direct link to Mihrr's post “Example 3 is bugged, yet ...” Example 3 is bugged, yet it has been bugged for a year apparantly. Fix it already. • (13 votes) redfoer 4 years agoPosted 4 years ago. Direct link to redfoer's post “Now example 3 does not ac...” Now example 3 does not accept the correct equation Keiko Whitlock a year agoPosted a year ago. Direct link to Keiko Whitlock's post “Well, I answered, y=-10x+...” Well, I answered, y=-10x+22, but it said wrong. But this is correct too, right? • (3 votes) Venkata a year agoPosted a year ago. Direct link to Venkata's post “Sign error. It should be ...” Sign error. It should be y = -10x - 22, not y=-10x + 22 (9 votes) Vector Inc. 3 years agoPosted 3 years ago. Direct link to Vector Inc.'s post “I don't understand step 2...” I don't understand step 2 of Example 2. How do they get the expanded version of (-1+h)^3? • (3 votes) ahmada timo 3 years agoPosted 3 years ago. Direct link to ahmada timo's post “(-1+h) x (-1+h) = get the...” (-1+h) x (-1+h) = get the result then multiply it in (-1+h) again (4 votes) Sbahle Thabede 7 months agoPosted 7 months ago. Direct link to Sbahle Thabede's post “can i have more question ...” can i have more question to practice • (4 votes) avs68325 9 months agoPosted 9 months ago. Direct link to avs68325's post “For question 1: what is s...” For question 1: what is step 4? How should I even interpret the question? This is so frustrating. For question 2: Why do we use the large formula? What is indicating that we have a g(t+h) function instead of a g(x)-g(t) function? And then I am asked to choose a point on the line after finding out what the slope of the tangent is. Is this darts? • (2 votes) Venkata 9 months agoPosted 9 months ago. Direct link to Venkata's post “Step 4 is just asking you...” Step 4 is just asking you to use all the information you already solved for (the slope and point) to get the equation of the tangent. You must already know that a unique line can be defined by a slope and a single point, so that's what they want you to do. You could use that formula too. With it, you'll get lim (x --> -1) (x^(3)+1)/(x+1). Expanding the numerator with the identity (a^(3) + b^(3) = (a+b)(a^(2) - ab + b^(2)), we get lim (x --> -1) (x+1)(x^(2) - 2x + 1)/(x+1). The x+1's cancel out, and you're left with lim (x --> -1) (x^(2) - 2x + 1). Substitute x = -1, and we get 3, which is exactly what we got with the other formula too. I reckon they used the other one because (a ± b)^(3) is more easy to calculate (incase you forget the identity) than a^(3) ± b^(3) (for which deriving the identity requires a but more effort) You could go the other way too, honestly. You can first find a point on the line and then find the slope. The exercise just wants to familiarize you with the idea and give an algorithm for students to follow. You are more than welcome to experiment and do stuff in a different way, as long as you obey all the rules of maths. (5 votes) sufyan.desai05 5 years agoPosted 5 years ago. Direct link to sufyan.desai05's post “In example 3, why does (-...” In example 3, why does (-5+h) get squared instead of (-5+3)? • (2 votes) Kiên P.S. 5 years agoPosted 5 years ago. Direct link to Kiên P.S.'s post “The function is f(x)=x^2+...” The function is f(x)=x^2+3 (For each value of x you gonna to square it then plus 3) (5 votes)Want to join the conversation?
y = -10x - 50 + 28
y = -10x - 22
And so for f(-5+h), you have to square (-5+h) first and then plus 3
Another way to say it, to calculate f(-5+h), whenever you see x in the function, you replace x with -5+h, and you'll get f(-5+h)=(-5+h)^2+3
That's why (-5+h) get squared
