for f(x) = 12-4x+2x^2, find an equation of the targent and normal at the point where x=-1 on curve with equation y=f(x)
2006-10-05 10:56:58 · 4 answers · asked by chefjordan 2 in Science & Mathematics ➔ Mathematics
y=- 8x + 10 , 8y - x - 145 = 0
2006-10-05 10:59:02 · answer #1 · answered by Anonymous · 0⤊ 0⤋
f(x)=12-4x+2x^2
f'(x)=-4+4x
at x=-1 f'(x)=-8
y=12+4+2=18
the equation of the tangent
y-18=-8(x+1)=>8x+y-10=0
and the normal slope=1/8
y-18=1/8(x+1)
so 8y-144=x+1 is the equation of the normal
=>x-8y+145=0
2006-10-05 11:11:30 · answer #2 · answered by raj 7 · 0⤊ 0⤋
f(x) = 12-4x+2x^2
f(-1) = 12-4(-1)+2(-1)^2 = 12+4+2 = 18
Tangent
f'(x) = -4+4x and f'(-1) = -4+4(-1) = -4-4 = -8
y=mx+c => 18=(-8)(-1)+c => c=18-8 => c=10
Equation of tangent: y = -8x + 10
Normal
m1xm2=-1 => m1x(-8)=-1 => m1=1/8
y=mx+c => 18=(1/8)(-1)+c => c=18+(1/8) => c=18 1/8
Equation of normal: y = (1/8)x + 18 1/8
2006-10-05 11:21:59 · answer #3 · answered by Kemmy 6 · 0⤊ 0⤋
at x=-1 from the curve...y=18
if y=2x^2-4x+12 differentiate wrt x
y'=4x-4 which at x=-1 is -8
this is the slope of tangent and that of normal obviously is then 1/8
eqn of tangent becomes using y=mx+c
8x+y=10 and so that of normal x-8y+145=0
2006-10-05 11:17:37 · answer #4 · answered by Anonymous · 0⤊ 0⤋