How do you find the equation of the tangent and normal line to the curve #y=x^2+2x+3# at x=1?
a. Equation of tangent,
b. Equation of normal,
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To find the equation of the tangent and normal lines to the curve (y = x^2 + 2x + 3) at (x = 1), follow these steps:
 Find the slope of the tangent line by taking the derivative of the curve with respect to (x), and then evaluate it at (x = 1).
 Use the slope obtained in step 1 and the point (P(1, f(1))) (where (f(x) = x^2 + 2x + 3)) to write the equation of the tangent line in pointslope form.
 Find the slope of the normal line by taking the negative reciprocal of the slope of the tangent line.
 Use the slope obtained in step 3 and the point (P(1, f(1))) to write the equation of the normal line in pointslope form.
Let's solve it step by step:
 Find the derivative of (y = x^2 + 2x + 3): [y' = 2x + 2]
Evaluate (y') at (x = 1): [y'(1) = 2(1) + 2 = 4]

Equation of the tangent line: [y  y_1 = m(x  x_1)] [y  (1^2 + 2(1) + 3) = 4(x  1)] [y  6 = 4(x  1)] [y  6 = 4x  4] [y = 4x + 2]

Slope of the normal line: [m_{\text{normal}} = \frac{1}{4}]

Equation of the normal line: [y  y_1 = m_{\text{normal}}(x  x_1)] [y  (1^2 + 2(1) + 3) = \frac{1}{4}(x  1)] [y  6 = \frac{1}{4}x + \frac{1}{4}] [y = \frac{1}{4}x + \frac{25}{4}]
So, the equation of the tangent line is (y = 4x + 2) and the equation of the normal line is (y = \frac{1}{4}x + \frac{25}{4}) at (x = 1).
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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.
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