How do you find the equation of the tangent and normal line to the curve #y=(2x+3)/(3x2)# at #(1,5)#?
Therefore, by the SlopePt. Form , the eqn. of tgt. is,
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To find the equation of the tangent and normal lines to the curve (y = \frac{{2x + 3}}{{3x  2}}) at the point (1, 5), follow these steps:
 Find the derivative of the given function.
 Evaluate the derivative at (x = 1) to find the slope of the tangent line.
 Use the pointslope form to write the equation of the tangent line.
 Determine the slope of the normal line by taking the negative reciprocal of the slope of the tangent line.
 Use the pointslope form to write the equation of the normal line.
Let's perform these steps:

Find the derivative of (y) with respect to (x): [ \frac{{dy}}{{dx}} = \frac{{d}}{{dx}}\left(\frac{{2x + 3}}{{3x  2}}\right) ] Using the quotient rule, we get: [ \frac{{dy}}{{dx}} = \frac{{(3x  2)(2)  (2x + 3)(3)}}{{(3x  2)^2}} ] [ \frac{{dy}}{{dx}} = \frac{{6x  4  6x  9}}{{(3x  2)^2}} ] [ \frac{{dy}}{{dx}} = \frac{{13}}{{(3x  2)^2}} ]

Evaluate the derivative at (x = 1): [ \frac{{dy}}{{dx}}\Bigg_{x=1} = \frac{{13}}{{(3(1)  2)^2}} = 13 ]

The slope of the tangent line at (x = 1) is (m = 13). Therefore, the equation of the tangent line using the pointslope form is: [ y  5 = 13(x  1) ]

The slope of the normal line is the negative reciprocal of the slope of the tangent line, so (m_{\text{normal}} = \frac{{1}}{{13}}).

Using the pointslope form, the equation of the normal line is: [ y  5 = \frac{{1}}{{13}}(x  1) ]
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To find the equation of the tangent line to a curve at a given point, you first need to find the derivative of the function, evaluate it at the given point to find the slope of the tangent line, and then use the pointslope form of a line to find the equation of the tangent line. Similarly, to find the equation of the normal line, you find the negative reciprocal of the slope of the tangent line and proceed with the pointslope form of a line using the given point.
First, find the derivative of the function ( y = \frac{2x+3}{3x2} ) using the quotient rule, then evaluate it at ( x = 1 ) to find the slope of the tangent line. After that, use the pointslope form of a line to find the equation of the tangent line. Finally, find the negative reciprocal of the slope of the tangent line to get the slope of the normal line, and use the pointslope form of a line again to find the equation of the normal line.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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