2006-08-15 11:22:16 · 4 answers · asked by jeep10455 2 in Education & Reference ➔ Higher Education (University +)
take the derivative...
y' = -10x
substitute the x value and get y' = -50
since y' is the slope
you now have the data for the point-slope form of a linear equation.....
y= m x + b
substituting the point
122 = -50 * 5 + b
122 = -250 + b
therefore b= 372
replacing the "b" value gives
y = -50x + 372
2006-08-15 11:32:39 · answer #1 · answered by Gemelli2 5 · 0⤊ 0⤋
well the slope of the tangent will be the derivative of the function so y' = -10x. Now we take the "x" coordinate of the point given, which is "5" and sub it in so y' = -10(5) = -50. Now -50 is the SLOPE of the tangent.
We also know a point (5,122) so simply :(122 - y) = -50(5 - x)
this gives us 122 - y = -250 + 50x, which can be re-arranged to :
y = -50x + 372 in SLOPE INTERCEPT FORM or
50x + y - 372 = 0 IN STANDARD form.
hope that helped
2006-08-15 20:37:03 · answer #2 · answered by jaymay2008 3 · 0⤊ 0⤋
no more math!!..i barely passed 1105 hehe
2006-08-15 18:27:30 · answer #3 · answered by icandaze 3 · 0⤊ 0⤋
In your pants?
2006-08-15 18:29:32 · answer #4 · answered by Alice Chaos 6 · 0⤊ 0⤋