+956
I asked a question like this about perpendicular lines, and I just wanted to clarify if I did this right following the same procedures for the perpendicular line.
f'(x) = 6x+1
6x+1=4
x = 2
f(2) = 14
(2,14)
Is that correct?
+129852 You have the correct idea....but a lttle mistake on the math
Derivative = 6x + 1
Set = to the slope of the line
6x + 1 = 4
6x = 3
x = 3/6 = 1/2
And y = 3(1/2)^2 + 1/2 = 5/4
So.....the point is ( 1/2, 5/4) and the equation of the line is
y = 4(x - 1/2) + 5/4
y = 4x - 2 + 5/4
y = 4x - 3/4
Here's the graph, Julius : https://www.desmos.com/calculator/48aijjgr6d
