14) Find the equations of the tangent and normal to the given curves at the indicated
points:
i) y = x^4 - 6x^3 + 13x^2 - 10x + 5 \: \: at\: \: (0, 5)

Answers (1)

We know that Slope of the tangent at a point on the given curve is given  by  \frac{dy}{dx}
Given the equation of the curve
y = x^4 - 6x^3 + 13x^2 - 10x + 5
\frac{dy}{dx}= 4x^3 - 18x^2 + 26x- 10
at point (0,5)
\frac{dy}{dx}= 4(0)^3 - 18(0)^2 + 26(0) - 10 = -10
Hence slope of tangent is -10
Now we know that,
slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{-10} = \frac{1}{10}
Now, equation of tangent at point (0,5) with slope = -10 is
y = mx + c\\ 5 = 0 + c\\ c = 5
equation of tangent is
y = -10x + 5\\ y + 10x = 5
Similarly, the equation of normal at point (0,5) with slope = 1/10 is
\\y = mx + c \\5 = 0 + c \\c = 5
equation of normal is
\\y = \frac{1}{10}x+5 \\ 10y - x = 50

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