at what points on the graph y=12x³+2x² does the tangent line have a slope of -2 ?
please explain it to me step by step!
if you find the answer you honestly would save my life
2007-01-25 11:55:13 · 4 answers · asked by libraaaalady 1 in Science & Mathematics ➔ Mathematics
The slope of a tangent is given by the value of the derivative at that point
y= 12x^3+2x^2 so y'= 36x^2 +4x This should be =-2
36x^2+4x+2=0 This equation has NO real roots so there are not points with a tangent of slope -2
Ther must be a mistake in your data but I outlined the proceeding
2007-01-25 12:15:36 · answer #1 · answered by santmann2002 7 · 0⤊ 0⤋
I'm going to help you out a bit...I don't have a calculator on hand and you'll have to do the work...
try graphing it and see if the slope is -2
Graph by inputting the x values 0,1,2,3,4,5,6...etc and solve for y until you get a decent graph
Eg. for x@1, y=12(1)^3+2(1)^2
x@2; y=12(2)^3+2(2)^2
and so on....
2007-01-25 20:10:20 · answer #2 · answered by CC 2 · 0⤊ 0⤋
step one
take the derivative
y'=36x^2+4x
step two set the equation equal to -2
36x^2+4x=-2
36x^2+4x+2=0
2(18x^2+2x+1)=0
and....theres no real roots so no solution
2007-01-25 20:05:12 · answer #3 · answered by aznskillz 2 · 0⤊ 0⤋
Set the derivative of y equal to -2 and solve for x.
2007-01-25 20:02:59 · answer #4 · answered by Professor Maddie 4 · 0⤊ 0⤋