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zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Hmm I think you've done a few of these types of problems. So I'm just curious, which result makes more sense to you. Seeing our tangent line in slopeintercept form: \(\large y=mx+b\), Or in pointslope form: \(\large yy_o=m(xx_o)\).

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0in slope intercept from

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Ok we will have 2 pieces we need to find. \(\large m\) is the slope of our tangent line. Take the derivative of your function, and evaluate it at \(\large x=\pi/8\). That will be your \(\large m\) value.

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0when i find the derivative a = pi/8 is not int he equation right?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Take the derivative of sin(4x) first. Don't plug a=pi/8 in until after you've taken a derivative.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1ok good, now plug in x=pi/8 :D

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0i got 0 after i plugged in and solved

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Ok sounds good. So we've determined that the slope \(\large m\) of our tangent line will be 0. \[\large y=mx+b \qquad \rightarrow \qquad y=0x+b\] Now we just need to find the \(\large b\) value.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1To find it, we'll need to plug in a coordinate pair that falls on the line. Well, since our line is tangent to the curve at x=pi/8, we can plug x=pi/8 into that function to get a corresponding y value. This will be the coordinate pair we'll plug into our tangent line.

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0so i should plug in pi/8 into y = mx + b?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1no, plug it into your original function, before you took the derivative of it.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1So that gives us a coordinate pair of \(\large \left(\dfrac{\pi}{8},\;1\right)\). Plug that into your tangent function to solve for \(\large b\).

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0ok so y = 0x + b so i should plug it in place of b?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1No, your coordinate pair is \(\large (x,\;y)\). Plug it into those.

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0okay so 1 = 0(pi/8) + b

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1So what is your b value? :3

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Ok sounds good. So if we plug that missing \(\large b\) into our tangent line equation, we get,\[\large y=0x+1\]Which can be simplified to,\[\large y=1\]

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1I guess it's really 3 steps. We had to find ~\(\large m\) ~\(\large b\) ~A Coordinate Pair.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1It's not too difficult, just a lot of silly letters :) Keep practicing!

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0okay so our equation is complete? and yes just lot to do and thanks

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, that's our final answer.
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