What is the equation of the tangent to y=5x^2-7x+4 at the point (2, 10)?

1 Answer
Nov 25, 2016

y=13x-16

Explanation:

The equation of the tangent is determined by finding the slope at
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the point x=2
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The slope is determined by differentiating y at x=2
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y=5x^2-7x+4
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y'=10x-7
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y'_(x=2) =10(2)-7

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y'_(x=2) =20 - 7=13
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The equation of the tangent of slope 13 and passing through the
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point (2,10) is:
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y-10=13(x-2)
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y-10=13x-26
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y=13x-26+10
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y=13x-16