The derivative tells you the slope at any point.
df/dx = 8 x^3 - 12 x
When x = 3, the slope is
m = 8*27 - 36 = 180
and the equation of the tangent line is
(y - 228) = 180*(x - 3)
y = 180 x - 540 + 228 = 180 x - 312
wow, that was absolutely correct (i just plugged it into webassign), thanks a lot!!
please just help me to understand the last part though for the test, where did you get the (y-228) = 180*(x-3)?
Find the equation of the line that is tangent to the graph of f(x) = 2x4 - 6x2 + 120 at the point (3, 228).
y=?
The equation for a line with slope m through points x = a and y = b is
(y- b)= m (x - a)
That is where the formula came from.
To find the equation of the tangent line at the point (3, 228), we can use the formula for a line with slope m through points x = a and y = b, which is:
(y - b) = m(x - a)
In this case, a = 3 and b = 228. We need to find the slope, which is the derivative of the function f(x). The derivative of f(x) = 2x^4 - 6x^2 + 120 is:
df/dx = 8x^3 - 12x
To find the slope at x = 3, substitute x = 3 into the derivative:
m = 8(3^3) - 12(3)
= 180
Now we have the slope, which is m = 180. Using the formula for the tangent line, we can substitute the slope and the point (3, 228):
(y - 228) = 180(x - 3)
This is the equation of the tangent line. Simplifying it further, we can write it in the standard form:
y - 228 = 180x - 540
y = 180x - 312
So the equation of the tangent line is y = 180x - 312.