How do you find the equation of a line tangent to the function y=(x-2)(x^2+1) at x=-1?

2 Answers
Apr 14, 2018

y=8x+2

Explanation:

Tangent Line to a Curve
simplify
y=x^3-2x^2+x-2
differentiate
y'=3x^2-4x+1
substitute for x=-1
y'=8
the value of y' is the slope of the tangent line at x=-1
and to get the point which lies on the tangent we substitute with x=-1 in the function
y=-6
so the point on the line is(-1,-6) and its slope is 8
substitute in the following formula to get the equation

y-y_1=m(x-x_1)

y+6=8(x+1)

y=8x+2

I hope this was helpful.

Apr 14, 2018

y = 8x + 2

Explanation:

At x = -1:

y = (-1-2)((-1)^2 + 1)

y = -3 xx 2

y = -6

The tangent passes through (-1,-6)

The gradient:

y = (x-2)(x^2 + 1)

y = x^3 - 2x^2 + x - 2

y' = 3x^2 - 4x + 1

At x = -1:

m = 3(-1)^2 - 4(-1) + 1

m = 3 + 4 + 1

m = 8

Since we have the gradient of the tangent and a point through which it passes we can use the point-gradient formula:

y - y_1 = m(x - x_1)

y +6 = 8(x +1 )

y = 8x + 8 - 6

y = 8x + 2