How do you find the equation of the tangent line to the curve #y = (3x-1)(2x+4)# at the point of (0,-4) ?

Question

Paisley Johnson · Accepted Answer

#y=10x-4#

#y=6x^2+10x-4#-> FOIL or multiply it out first #y'=12x+10#-> find the derivative #y'(0)=10#->Put in 0 for x to find slope #y+4=10(x-0)# ->Put the slope #m# and the point# (x_1,y_1)# into the point slope form of an equation of a line #y=10x-4#-Write the answer in slope-intercept form

Note you can start by finding the derivative using the product rule whatever is easier for you go with it.

Scarlett Allison · Answer

undefined To find the equation of the tangent line to the curve y = (3x-1)(2x+4) at the point (0,-4), we need to find the slope of the tangent line at that point.

First, we find the derivative of the function y = (3x-1)(2x+4) using the product rule.

The derivative is given by: y' = (3x-1)(2) + (2x+4)(3)

Simplifying this expression, we get: y' = 6x - 2 + 6x + 12

Combining like terms, we have: y' = 12x + 10

Now, we substitute x = 0 into the derivative to find the slope at the point (0,-4).

Substituting x = 0 into y', we get: y' = 12(0) + 10 = 10

Therefore, the slope of the tangent line at the point (0,-4) is 10.

Using the point-slope form of a linear equation, we can write the equation of the tangent line as:

y - y1 = m(x - x1)

Substituting the values of the point (0,-4) and the slope m = 10, we have:

y - (-4) = 10(x - 0)

Simplifying this equation, we get:

y + 4 = 10x

Rearranging the equation to the standard form, we have:

10x - y = -4

Therefore, the equation of the tangent line to the curve y = (3x-1)(2x+4) at the point (0,-4) is 10x - y = -4.