My equation came out to be
Y=22x-120....and it was not correct....what did I do wrong?
2007-11-14 16:58:35 · 5 answers · asked by RedSparkle 1 in Science & Mathematics ➔ Mathematics
f(x) = 2x^2 -10x + 4x - 20
= 2x^2 - 6x - 20
f'(x) = 4x - 6
f'(7) = 28 - 6 = 22
Since the slope is 22, the y intercept is when x = 0
You know one point on the line is at (7,36) and with a slope of 22 (*7) the y intercept of the tangent line will be 154 below that point or -118 - the official way is to use the point slope formula.
y-y1 = (x-x1)m
y-36 = ((0)-7)(22)
y-36 = -154
y = -118
y=22x-118
You were really close!!
2007-11-14 17:13:23 · answer #1 · answered by Mark P 2 · 0⤊ 0⤋
2x+4+2x-10 is the derivative
dy/dx=4x-6
plug in 7...
28-6
22 is the slope
y=22x+b
(7-5)(14+4)=22(7)+b
2(18)-154=118
y=22x-118 is the answer
2007-11-14 17:03:53 · answer #2 · answered by mathwhiz620 2 · 0⤊ 0⤋
f'(x) = 2x+4 + 2(x+-5)
f'(x) = 2x+4 + 2x +-10
f'(x) = 4x + -6
plug in 7 to find m
4(7) + -6 = 22
y=22x+b
(7-5)(2(7)+4) = 22(7)+b
36 = 154 + b
b= -118
y= 22x - 118
2007-11-14 17:05:50 · answer #3 · answered by Matty B 3 · 0⤊ 0⤋
Back to basics. You have to multiply out the factors to get the quadratic. Once you do that, the SLOPE of the tangent line is 2ax + b, where a is the quadratic term coefficent and b is the x coefficient. Find this for x=7. Then back to the quadratic to find the y value for x=7. Then if the tangent line is Y= SLOPE x + Intercept, you substitute in your value of (x,y) to find the intercept.
2007-11-14 17:05:22 · answer #4 · answered by cattbarf 7 · 0⤊ 0⤋
its just a subtraction mistake.
its y= 22x-118
not 120 :D
2007-11-14 17:05:15 · answer #5 · answered by winnydamarpoe 2 · 0⤊ 0⤋