How do you find the equation of the tangent and normal line to the curve #y=x^2+2x+3# at x=1?

Answer 1

a. Equation of tangent, # y =4 x + 2#
b. Equation of normal, #4 y = -x +25#

#y = x^2 + 2 x + 3#
at #x =1#, #y = 1^2+ 2(1) + 3 =6#
let say #m_1# = gradient of tangent. #(d y)/(d x) = 2 x + 2#
at #x =1#, #m_1 = (d y)/(d x) = 2 (1) + 2 =4#
Therefore the equation of tangent at (1,6), #(y-6) = m_1(x-1)# #(y-6) = 4(x-1)# # y =4 x -4 +6# # y =4 x + 2#
let say #m_2# = gradient of normal. #m_1 * m_2 = -1# #m_2 = -1/4#
Therefore the equation of normal at (1,6), #(y-6) = m_2(x-1)# #(y-6) = -1/4(x-1)#
#y = -1/4 x + 1/4 +6#
#y = -1/4 x + 25/4#
#4 y = -x +25#
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Answer 2

To find the equation of the tangent and normal lines to the curve (y = x^2 + 2x + 3) at (x = 1), follow these steps:

  1. 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).
  2. 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 point-slope form.
  3. Find the slope of the normal line by taking the negative reciprocal of the slope of the tangent line.
  4. Use the slope obtained in step 3 and the point (P(1, f(1))) to write the equation of the normal line in point-slope form.

Let's solve it step by step:

  1. Find the derivative of (y = x^2 + 2x + 3): [y' = 2x + 2]

Evaluate (y') at (x = 1): [y'(1) = 2(1) + 2 = 4]

  1. 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]

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

  3. 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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Answer from HIX Tutor

When evaluating a one-sided 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 one-sided 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 one-sided 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 one-sided 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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