What is the equation of the line that

Ariel Collier

Ariel Collier

Answered question

2022-04-08

What is the equation of the line that is normal to f(x)=2x22ex at x=-2?

Answer & Explanation

rhyclelal80j6

rhyclelal80j6

Beginner2022-04-09Added 13 answers

f(x)=2x22ex
f(x)=4x2ex
f(2)=82e2 This is the gradient of the tangent.
mn=1mtan=182e2=e28e22
f(2)=82e2=8e22e2
yy1=m(xx1)
y+8e22e2=e28e22(x+2)
y+8e22e2=xe28e222e28e22
y=xe218e233e416e2+2e2(2e1)(2e+1)
muthe2ulj

muthe2ulj

Beginner2022-04-10Added 10 answers

Given: f(x)=2x22ex and x0=2
y0=f(2)=82e2
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)=(2x22ex)=2(2x+ex)=2(2x+ex)
Hence, M(x0)=1f(x0)=12(2x0+ex0)
Next, find the slope at the given point.
m=M(2)=12(4+e2)
Finally, the equation of the normal line is yy0=m(xx0)
Plugging the found values, we get that y(82e2)=12(4+e2)(x(2))
Or, more simply: y=x2(4+e2)+233e4e2+4e4

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