How do you find the equation of the line tangent to #y=4x^3+12x^2+9x+7# at (-3/2,7)?

Question

Christopher Agresta · Accepted Answer

#y=7#

The slope (gradient) of the line is the rate of change in y for the rate of change in x

So we have #" "("change in y")/("change in x") =(dy)/(dx)#
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Let the gradient of the line be #m#
Let any point on the line be P

Given: #" "y=4x^3+12x^2+9x+7#

From this:#" "m=(dy)/(dx) = 12x^2+ 24x+9#
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At the point #(x,y)->(-3/2,7)#

#m=(dy)/(dx)=12(-3/2)^2+24(-3/2)+9 #

#m= +27-36+9 = 0 #
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So the equation of the tangent is#" " y=(0)x+c#

This tangential line passes through the point #P_(-3/2,7)#

So we have:

#7=(0)(-3/2)+c#

#c=7#
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So the equation of the tangent at #P_(-3/2,7)#

is: #" "y=+7#
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Charles Anctil · Answer

undefined To find the equation of the line tangent to the curve y=4x^3+12x^2+9x+7 at the point (-3/2,7), we need to find the slope of the tangent line at that point.

First, we find the derivative of the given function y=4x^3+12x^2+9x+7. The derivative is dy/dx = 12x^2 + 24x + 9.

Next, we substitute the x-coordinate of the given point (-3/2,7) into the derivative to find the slope at that point.

dy/dx = 12(-3/2)^2 + 24(-3/2) + 9 = 54/4 - 36/2 + 9 = 27/2 - 36/2 + 9 = 0.

Since the slope is 0, the equation of the tangent line is y = 7.