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Find the equation of the tangent to the curve y=2x^2+3 that pass through point (2,-7)
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You need to use the formula that gives the equation of the tangent line to a curve, at a point,`(x_0,y_0)` , such that:
`f(x) - f(x_0) = f'(x_0)(x - x_0)`
The problem provides the coordinates `(x_0,f(x_0)) = (2,-7)` and you need to evaluate `f'(x)` at `x_0 = 2` , such that:
`f'(x) = 4x => f'(2) = 4*2 => f'(2) = 8`
`f(x) - (-7) = 8(x - 2) => f(x) + 7 = 8x - 16`
Substituting y for f(x) yields:
`8x - y - 16 - 7 = 0 => 8x - y - 23 = 0`
Hence, evaluating the equation of the tangent line to the curve, at the point `(2,-7)` yields `8x - y - 23 = 0.`
Posted by sciencesolve on March 7, 2013 at 5:44 PM (Answer #1)
Honors, Dean's List
y=`2x^2 + 3`
gradient = 4(2)
`Let x_(o) = 2 and y_(o) =-7`
Posted by jacqueshaw on March 8, 2013 at 5:33 AM (Answer #2)
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