What is the equation of the normal line of f(x)=x−3x+4 at x=3?

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Wernbergbo9d · Accepted Answer

Explanation:As long as f′(x)≠0, the normal line of f(x) at x can be written on the formy=ax+m, where a is the slope and m is the intercept with the y-axis (see further down what happens for the case f&#039;(x)=0). If we first find a, then we can find m through elimination.We know that the normal line of f(x) is perpendicular to the tangent line of f(x) at x=3. Therefore, if we can find the slope of f(x) at x=3, then we can also find the slope of the normal line at x=3.Denote the slope of the tangent line b. If a,b≠0, then a*b=-1 (see further down for explanation).The slope of the tangent line at a value x is by definition f&#039;(x), which we can compute using the product rule (or quotient rule) .f′(x)=1x+4−x−3(x+4)2=x+4(x+4)2−x−3(x+4)2=(x+4)−(x−3)(x+4)2=7(x+4)2.Therefore b=f′(3)=772=17, and we find the slope of the normal line by solving a*b=-1 for a, which gives thata=-7.Since the normal line passes through x=3, we know that the point (3, f(3)) must lie on the line. Evaluating f(3)=0, and inserting in the equation for the normal line we get that0=a*3+mInserting a=-7 and solving for m gives thatm=21,and we have found that the normal line is given byy=-7x+21.

What is the equation of the normal line

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