How do you give the equation of the normal line to the graph of #y=2xsqrt(x^2+8)+2# at point (0,2)?

Answer 1
The answer is: #y=-sqrt2/8x+2#.

The normal line to the graph in one point is the perpendicular at the tangent line to the graph in that point.

Remembering that two lines #n# and #t# are perpendicular if and only if this rule is verified:
#m_n=-1/m_t# , where #m# is the slope,

and

the slope of the tangent line in a point to a curve is the first derivative in that point,

#y'=2(1*sqrt(x^2+8)+x*1/sqrt(x^2+8)*2x)#

and so:

#y'(0)=2sqrt8=4sqrt2#.

So:

#m_t=4sqrt2rArrm_n=-1/(4sqrt2)=-1/(4sqrt2)*sqrt2/sqrt2=-sqrt2/8#.
The line that passes from a given a point #P(x_P,y_P)# and with a slope #m#, is:
#y-y_P=m(x-x_P)#

so:

#y-2=-sqrt2/8(x-0)rArry=-sqrt2/8x+2#.
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Answer 2

To find the equation of the normal line to the graph of y=2xsqrt(x^2+8)+2 at point (0,2), we need to determine the slope of the tangent line at that point.

First, we find the derivative of the given function y=2xsqrt(x^2+8)+2 with respect to x.

Using the product rule and chain rule, the derivative is:

dy/dx = 2sqrt(x^2+8) + 2x(1/2)(x^2+8)^(-1/2)(2x)

Simplifying further, we have:

dy/dx = 2sqrt(x^2+8) + 2x^2/(sqrt(x^2+8))

Next, we substitute x=0 into the derivative to find the slope of the tangent line at (0,2):

dy/dx = 2sqrt(0^2+8) + 2(0)^2/(sqrt(0^2+8))

Simplifying, we get:

dy/dx = 2sqrt(8) + 0

Therefore, the slope of the tangent line at (0,2) is 2sqrt(8).

Since the normal line is perpendicular to the tangent line, the slope of the normal line is the negative reciprocal of the slope of the tangent line.

Hence, the slope of the normal line is -1/(2sqrt(8)).

Using the point-slope form of a line, we can write the equation of the normal line:

y - 2 = (-1/(2sqrt(8)))(x - 0)

Simplifying, we have:

y - 2 = (-1/(2sqrt(8)))x

Rearranging the equation, we get the equation of the normal line to the graph of y=2xsqrt(x^2+8)+2 at point (0,2):

y = (-1/(2sqrt(8)))x + 2

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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