Asked by drwls
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.
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.
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