- Can you explain to me how you got it cause i would like to understand (: Thanks!
Find the points on the curve y = 2x^3 + 3x^2 − 12x + 7where the tangent line is horizontal.
(x,y)= (_____) (smaller x-value)
(x,y)= (_____) (larger x-value)
Differentiate the function.
g(x) = x^2(1 − 5x)
Find the points on the curve y = 2x^3 + 3x^2 − 12x + 7where the tangent line is horizontal.
(x,y)= (_____) (smaller x-value)
(x,y)= (_____) (larger x-value)
Differentiate the function.
g(x) = x^2(1 − 5x)
-
y = 2x^3 + 3x^2 − 12x + 7
to find the tangent line, you need to find its slope by taking derivative of y
y' = 6x²+6x -12
and since the slope of a horizontal line = 0
let y' =0, solve for x
plug value of x back into y
since y' is quadratic, you will have two solutions.
to find the tangent line, you need to find its slope by taking derivative of y
y' = 6x²+6x -12
and since the slope of a horizontal line = 0
let y' =0, solve for x
plug value of x back into y
since y' is quadratic, you will have two solutions.