Find the equation of the tangent and normal lines to the functiony = x2 - 6x + 4...

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Ronsmart @Ronsmart

September 2022 1 0 Report

Find the equation of the tangent and normal lines to the function
y = x2 - 6x + 4 at the point (2,-4).

Answers & Comments

$\mathbb{SOLUTION:}$

To determine the slope of the tangent line, get the first derivative of the curve and evaluate it at the given x-coordinate of the point.

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$\mathsf{\dfrac{d}{dx}(y=x^2-6x+4)}$

$\mathsf{\dfrac{dy}{dx}=2x-6} \longleftarrow \textsf{evaluate at x = 2}$

$\mathsf{\dfrac{dy}{dx}=2(2)-6}$

$\mathsf{\dfrac{dy}{dx}=-2}$

$></p><p>The slope of the tangent line is m₁ = -2, the tangent line is also passing through point (2, -4).</p><p>Using point-slope form of a line:</p><p><img src=$

$\mathsf{m=m_1=-2}$

$\mathsf{(x_1,y_1)=(2,-4)}$

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$\mathsf{y+4=-2(x-2)}$

$\mathsf{y+4=-2x+4}$

$\mathsf{y=-2x} \Longleftarrow \textsf{equation of tangent line}$

$></p><p>The slope of the normal line is the negative reciprocal of the slope of the tangent line. m₂ = -(1/m₁). The normal line is also passing through point (2, -4).</p><p><img src=$

$\mathsf{m_2=\dfrac{1}{2}}$

Using point-slope form of a line:

$\mathsf{y-y_1=m(x-x_1)}$

$\mathsf{m=m_2=\dfrac{1}{2}}$

$\mathsf{(x_1,y_1)=(2,-4)}$

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$\mathsf{y+4=\dfrac{1}{2}x-1}$

$\mathsf{y=\dfrac{1}{2}x-5} \Longleftarrow \textsf{equation of the normal line}$

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The equation of the tangent line is: $\boxed{\textsf{y = -2x}}$

The equation of the normal line is: $\boxed{\mathsf{y=\dfrac{1}{2}x-5}}$

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