Answer:
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 ) = 4 x 3 + 12 x 2 + 9 x + 7
= f ( x ) = 4 x 3 + 12 x 2 + 9 x 1 + 7 x 0
Using the power rule, let's now compute the derivative of f( x )
f ' ( x ) = ( 3 ⋅ 4 x 2 ) + ( 2 ⋅ 12 x ) + ( 1 ⋅ 9 x 0 ) + 0 ⋅ 7 x − 1
f ' ( x ) = 12 x 2 + 24 x + 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:
y=45x+115