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Sagot :
The equations of the lines in the form y=mx+b are-
- Tangent line: y = 240x - 215994
- Normal line: y = - 0.00417x + 9.75
What is slope?
A line's slope is described like the change in y coordinate to the change in x coordinate. The net y coordinate change is y, whereas the net x coordinate change is x.
Therefore the variation in y coordinate in relation to the variation in x coordinate is written as, dy/dx.
Now, as per the given conditions;
The equation of the line is y = (6 + 4x)²
Simplifying the equation;
y = (6 + 4x)²
= 36 + 48x + 16x²
y = 16x² + 48x + 36
Now, differentiate the equation to find the tangent slope;
dy/dx = 32x + 48
At the point (6,900),
dy/dx = 32(6) + 48 = 240
Let equation of the tangent at point (a,b) is;
(y - b) = m(x - a)
a = 6, b = 900, m = 240
y - 6 = 240(x - 900)
Write this in y = mx + b form,
y - 6 = 240x - 216000
y = 240x - 215994 (Equation of Tangent line)
The slope of the normal line = -(1/slope of the tangent line) (since they're both perpendicular to each other)
Slope of the normal line = -1/240
The, equation of the normal will become;
y - 6 = (-1/240)(x - 900)
y - 6 = (-x/240) + 3.75
y = (-1/240)x + 9.75
y = - 0.00417x + 9.75 (Equation of Normal line)
Thus, the equation for the normal and tangent has be found.
To know more about the slope of the line, here
https://brainly.com/question/16949303
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The complete question is-
Find the equation of the tangent line and normal line to the curve y = (6 + 4x)² at the point (6,900). Write the equations of the lines in the form y=mx+b. Tangent line: y=
Normal line: y=
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