How do you find the equation of the tangent to the curve with equation #y = x^(3) + 2x^(2) - 3x + 2# at the point where x = 1?

Question

Isaiah Allen · Accepted Answer

Tangent has equation: #y=4x-2#

First we calculate #f(1)# to find the point at which tangent and the curve meet.

#f(1)=1+2-3+2=2#

So we are looking for a tangent in #(1,2)#.

The tangent line has equation of: #y=ax+b#, where #a=f'(1)#, so we have to calculate #f'(1)#.

#f'(x)=3x^2+4x-3# #f'(1)=3+4-3=4#

So the tangent line has equation #y=4x+b#, now we are looking for such #b#, for which the tangent line passes through #(1,2)#

#2=4+b#

#b=-2#

Finally the tangent line has equation of #y=4x-2#

graph{(y-x^3-2x^2+3x-2)(y-4x+2)=0 [-17.1, 11.37, -3.28, 10.96]}

Axel Alvarez · Answer

undefined To find the equation of the tangent to the curve at the point where x = 1, we need to find the slope of the tangent line and the coordinates of the point of tangency.

First, we find the derivative of the given equation to find the slope of the tangent line. The derivative of y = x^3 + 2x^2 - 3x + 2 is dy/dx = 3x^2 + 4x - 3.

Next, we substitute x = 1 into the derivative to find the slope at that point. dy/dx = 3(1)^2 + 4(1) - 3 = 4.

Now, we need to find the y-coordinate of the point of tangency. We substitute x = 1 into the original equation y = x^3 + 2x^2 - 3x + 2. y = (1)^3 + 2(1)^2 - 3(1) + 2 = 2.

Therefore, the point of tangency is (1, 2) and the slope of the tangent line is 4.

Using the point-slope form of a linear equation, y - y1 = m(x - x1), we can substitute the values we found to get the equation of the tangent line.

y - 2 = 4(x - 1)

Simplifying, we get y = 4x - 2 as the equation of the tangent to the curve at the point where x = 1.