How do you find the equation of a line tangent to the function $y = x^{3} - 2 x^{2} + 2$ $y=x^3-2x^2+2$ at (2,2)?

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How do you find the equation of a line tangent to the function $y = x^{3} - 2 x^{2} + 2$ $y=x^3-2x^2+2$ at (2,2)?

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Answer 1 · 2017-05-19T02:04:08

$y = 4 x - 6$ $y = 4x - 6$

Explanation:

First, find the derivative of $y$ $y$ , which is

$y = 3 x^{2} - 4 x$ $y = 3x^2 - 4x$

Now, plug in the $x$ $x$ -value ( $2$ $2$ ) into this equation to find the slope of the tangent line:

$y = 3 {(2)}^{2} - 4 (2) = 4$ $y = 3(2)^2 - 4(2) = 4$

Now plug the $x$ $x$ -value (2) back into the original equation to find the $y$ $y$ -coordinate of the tangential point:

$y = {(2)}^{3} - 2 {(2)}^{2} + 2 = 2$ $y = (2)^3 - 2(2)^2 + 2 = 2$

(Technically I didn't need to do this since the $y$ $y$ -coordinate was given; I did it anyway so you could find the tangent line with just a given $x$ $x$ -coordinate.)

Lastly, use the point-slope formula to find the equation of the line tangent to the function at $x = 2$ $x=2$ :

$y - y_{1} = m (x - x_{1})$ $y-y_1 = m(x-x_1)$

$y - 2 = 4 (x - 2)$ $y - 2= 4(x-2)$

$y = 4 x - 6$ $y = 4x - 6$

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How do you find the equation of a line tangent to the function $y = x^{3} - 2 x^{2} + 2$ $y=x^3-2x^2+2$ at (2,2)?

1 Answer

Explanation:

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How do you find the equation of a line tangent to the function $y = x^{3} - 2 x^{2} + 2$ $y=x^3-2x^2+2$ at (2,2)?