Question

I am trying to find an equation of the tangent to the curve y = 4 sin^2 x at the point (π/6,1). I already figured out that y' = 2√3. Any suggestions?

Answer 1

So you know that is the gradient, now just use the formula
y-y1=m(x-x1)
And input the coordinates of the points and you have your equation.

Answer 2

Awesome! I know that equation but what is it called again?

Answer 3

Oh, I see. You just rearranged the slope equation: \[m = (y _{2}-y _{1})/(x _{2}-x _{1})\]

Answer 4

Thank you!

Answer 5

that equation is called Slope - Point" form of equation of line.

Answer 6

ofcourse because when we know slope and a point on a line, then we can use it.

Answer 7

Ahh, thank you!