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Sagot :
The equation of the line normal to the curve y = square root(3x² + 2x) at point (2,4) is given as follows:
y - 4 = -4/7(x - 2).
How to obtain the equation of the line normal to the curve?
The equation of the line normal to a curve follows the point-slope definition of a linear function, given as follows:
y - y' = m(x - x').
The function in this problem is defined as follows:
y = square root(3x² + 2x).
Then the derivative of the function, applying the chain rule, is given as follows:
y' = (6x + 2)/(2 x square root(3x² + 2x))
y' = (3x + 1)/square root(3x² + 2x)
At x = 2, the numeric value of the derivative is of:
y' = 7/square root(16) = 7/4.
The slope of the normal line is calculated as follows:
m x 7/4 = -1
m = -4/7.
(as the normal line is perpendicular to the tangent line, which has slope equals to the numeric value of the derivative at the point).
The coordinates of the point at (2,4), hence:
x' = 2, y' = 4.
Then the definition of the normal line is given as follows:
y - 4 = -4/7(x - 2).
More can be learned about linear functions at https://brainly.com/question/24808124
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