dy/dx=-2=6x^2+4x-9
6x^2+4x-7=0
x=(-4+-sqrt(16+4*6*7))/12
x=(-4+-13.6)/12
those are the two points. Check my math with a calc, I did it in my head.
At what points on the graph of f(x)=6x^2+4x-9 is the slope of the tangent line -2?
2 answers
Bob forgot to find the derivative first
dy/dx = 12x + 4
12x+4 = -2
12x = -8
x = -8/12 = -2/3
f(-2/3) = 6(4/9) + 4(-2/3) - 9 = -25/3
We have a parabola shown below with a tangent at (-2/3 , -25/3) having a slope of -2
http://www.wolframalpha.com/input/?i=plot+y%3D6x%5E2%2B4x-9
dy/dx = 12x + 4
12x+4 = -2
12x = -8
x = -8/12 = -2/3
f(-2/3) = 6(4/9) + 4(-2/3) - 9 = -25/3
We have a parabola shown below with a tangent at (-2/3 , -25/3) having a slope of -2
http://www.wolframalpha.com/input/?i=plot+y%3D6x%5E2%2B4x-9
3 answers