Question

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

Answer 1

`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)`