Question 181349: Determine the equations of both lines that are tangent to the graph of f(x)=x^2 and pass through point (1,-3).
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Determine the equations of both lines that are tangent to the graph of f(x)=x^2 and pass through point (1,-3).
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The slope of f(x) at ever x is 2x
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You want the equation of the line thru (x,x^2) and (1,-3) and has slope = 2x.
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slope = (x^2--3)/(x-1) = 2x
(x^2+3) = 2x(x-1)
x^2 + 3 = 2x^2 - 2x
x^2 - 2x -3 = 0
(x-3)(x+1) = 0
x = -1 or x = 3
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The tangent points on the curve are (-1,1) and (3,9)
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Equation of the tangent passing thru (-1,1), (1,-3)
slope = (-3-1)/(1--1) = -4/2 = -2
intercept?:
1 = -2*-1 + b
b = -1
Equation: y = -2x - 1
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Equation of the tangent passing thru (3,9) and (1,-3)
slope = (9--3)/(3-1) = 6
-3 = 6*1 + b
b = -9
Equation: y = 6x - 9
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Cheers
Stan H.
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