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Find the equation to the tangent at the curve

2007-03-01 00:42:43 · 4 answers · asked by stuey11 1 in Science & Mathematics Mathematics

4 answers

gradient at tangent:
dY/dx= 2x = 2(3)=6 (at point [3,0])

since the tangent is a straight line: y=mx+c m:gradient=6 and c:constant/ y-intercept

=> y=6x+c___________(i)
when x=3, y=0 (substitute this into (i))
0=6(3)+c
=> c=-18

therefore equation to the tangent at the curve is:
y=6x-18

2007-03-01 00:49:54 · answer #1 · answered by sadia1905 3 · 0 0

I think you want to say: "Find the equation OF the tangent line to the curve: Y = x^2 - 9 @ the point (3,0)"

...in which case Sadia has the line equation you want:

y = 6x - 18

Alex... seems to have misread your typing AND SO DID I.
Alex has covered both possibilities.

2007-03-01 09:04:18 · answer #2 · answered by answerING 6 · 0 0

For case y = 2*x - 9
dy/dx = 2
equation of line is
y-0 = 2(x-3)
y = 2x-6

For case y = x^2 -9
dy/dx = 2x
at x = 3, dy/dx = 6
equation of line is
y - 0 = 6(x-3)
y = 6x-18

2007-03-01 08:51:24 · answer #3 · answered by AlexTan 3 · 0 0

Differentiate it.
dy/dx = 2x, = 2(3), = 6

2007-03-01 08:56:44 · answer #4 · answered by nayanmange 4 · 0 0

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