Find the equation of the tangent to the curve y=x^3-6x^2+7x+3 at (-1,1).

To find the equation of the tangent line to the curve at the point (-1,1), we first need to find the slope of the tangent line at that point.

The slope of the tangent line to a curve at a given point is equal to the derivative of the function at that point. So, we need to find the derivative of the function y = x^3 - 6x^2 + 7x + 3.

y' = 3x^2 - 12x + 7

Now, we can find the slope of the tangent line at the point (-1,1) by plugging in x = -1 into the derivative:

y'(-1) = 3(-1)^2 - 12(-1) + 7
y'(-1) = 3 + 12 + 7
y'(-1) = 22

So, the slope of the tangent line at the point (-1,1) is 22.

Now that we have the slope of the tangent line and a point on the line, we can use the point-slope form of a linear equation to find the equation of the tangent line:

y - y1 = m(x - x1)

Plugging in the values we have:

y - 1 = 22(x - (-1))
y - 1 = 22(x + 1)
y - 1 = 22x + 22
y = 22x + 23

Therefore, the equation of the tangent line to the curve y = x^3 - 6x^2 + 7x + 3 at the point (-1,1) is y = 22x + 23.

The equation of the tangent to a curve at a point is given by:

$$y - y_1 = m(x - x_1)$$

where $(x_1, y_1)$ is the point of tangency and m is the slope of the tangent.

To find the slope of the tangent, we need to find the derivative of the curve at the point of tangency.

$$f'(x) = \frac{d}{dx}(x^3-6x^2 7x 3)$$

$$= 3x^2-12x+7$$

Evaluating the derivative at the point of tangency (-1,1) gives us the slope of the tangent:

$$f'(-1) = 3(-1)^2-12(-1)+7 = 16$$

Now we can plug the values of $x_1$, $y_1$, and m into the equation of the tangent to get the equation of the tangent to the curve y=x^3-6x^2 7x 3 at (-1,1):

$$y - 1 = 16(x - (-1))$$

$$y - 1 = 16x + 16$$

$$y = 16x + 17$$