What is the equation of the line that is normal to f(x)=x3−5x+2 at x=3?

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Egerlandsq0z · Accepted Answer

Explanation:Take the derivative of f(x). The derivative of f(x) isf′(x)=3(x2)−5Plug 3 into the derivative to getf′(3)=3(32)−5=22That is the slope tangent to the curve but to find the slope of the tangent line, you need to take the reciprocal of that number and then switch signs, so the slope of the normal line is −122.Now plug 3 into the initial equation.f(3)=33−5(3)+2=14Now that we have the point (3, 14) and the slope of the normal line, we can plug into the equation(y−y1)=dydx(x−x1)Plug in 14 for y1,−122 for dydx, and 3 for x1. Solve for y and you should gety=(−x22)+(4522)

Jax Burns · Accepted Answer

We are given that f(x)=x3−5x+2 and x0=3.Find the value of the function at the given point: y0=f(3)=14.The slope of the normal line at x=x0 is the negative reciprocal of the derivative of the function, evaluated at x=x0:M(x0)=−1f′(x0)Find the derivative: f′(x)=(x3−5x+2)′=3x2−5Hence, M(x0)=−1f′(x0)=−13x02−5Next, find the slope at the given point.m=M(3)=−122Finally, the equation of the normal line is y−y0=m(x−x0)Plugging the found values, we get that y−14=−x−322Or, more simply: y=31122−x22.

What is the equation of the line that

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