Question

16. Find the equation of tangent and normal to the curve at the indicated point y = x² - 6x³ + 13x² - 10x + 5 at (0, 5).
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Answer 1

Resposta:

Let's simplify the given function:

$y = x^{2} - 6x^{3} + 13x^{2} - 10x + 5\\y = -6x^{3} +14x^{2} - 10x + 5$

The slope of the tangent line at a point on the function is equal to the derivative of the function at the same point. Thus:

$y' = -18x^{2}+28x - 10\\y'(0) = -10$

The line intercept on the y-axis is 5. Thus, the equation of the tangent line is:

$y = mx + b\\y = -10x + 5$

From analytic geometry, the slope of any line perpendicular to a line with slope m is the negative reciprocal −1/m. Thus, the equation of the normal line is:

$y = mx + b\\y = \frac{1}{10}x + 5$