What is the equation of the line tangent to #f(x)=4x^3+12x^2+9x+7# at #x=-3#?

1 Answer
Apr 11, 2018

#color(purple)(y=45x+115)#

Explanation:

We can find the slope of a line tangent to a curve at a point by evaluating the derivative of the function at that point.

We are given the function

#f(x) = 4x^3 + 12x^2 + 9x + 7#

#=f(x) = 4x^3 + 12x^2 + 9x^1 + 7x^0#

Using the power rule, let's now compute the derivative of #f(x)#

#f'(x) = (3*4x^2)+(2*12x)+(1*9x^0)+cancel(0*7x^-1)#

#f'(x) = 12x^2+24x+9#

We can now find the slope of #f(x)# at #x=-3# by substituting this value into #f'(x)#

#f'(-3)=12(-3)^2+24(-3)+9#

#f'(-3)=12(9)+24(-3)+9#

#f'(-3)=108-72+9#

#f'(-3)=45# (slope of the tangent line at #x=-3#

Now that we have a slope for the tangent line, we need to identify a point on the line.

We know the tangent line touches the function #f(x)# at the point #x=-3#, so let's find the value of #f(x)# at this point:

#f(-3) = 4(-3)^3 + 12(-3)^2 + 9(-3) + 7#

#f(-3) = 4(-27) + 12(9) + 9(-3) + 7#

#f(-3) = -108 + 108 -27 + 7#

#f(-3) = -20#

So we know the tangent line goes through the point

#(-3 , -20)#

Finally, we can use the point-slope formula for a line to find the equation of the tangent line.

#y=mx+b#

To find the value of #b#, substitute the values we have calculated for the point and slope of the tangent line:

#(-20)=(45)(-3)+b#

#-20=-135+b#

#b=115#

So our final answer for the equation of the tangent line is:

#color(purple)(y=45x+115)#