Slove using English

Question

Slove using English

ScanSolve AI Tutor · Accepted Answer

Here's the solution to finding the equation of the tangent to the curve at the given point:

The curve is given by the equation $y = x^3 - 3x^2 + 2x - 1$ .
The point of tangency is $(3, 5)$ .

Find the first derivative of the curve to determine the slope of the tangent line.
$y = x^3 - 3x^2 + 2x - 1$ $\frac{dy}{dx} = \frac{d}{dx}(x^3) - \frac{d}{dx}(3x^2) + \frac{d}{dx}(2x) - \frac{d}{dx}(1)$ $\frac{dy}{dx} = 3x^2 - 3(2x) + 2(1) - 0$ $\frac{dy}{dx} = 3x^2 - 6x + 2$
Evaluate the derivative at the given x-coordinate of the point of tangency, $x=3$ , to find the slope $m$ .
$m = \left.\frac{dy}{dx}\right|_{x=3} = 3(3)^2 - 6(3) + 2$ $m = 3(9) - 18 + 2$ $m = 27 - 18 + 2$ $m = 9 + 2$ $m = 11$
Use the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$ , with the point $(x_1, y_1) = (3, 5)$ and the slope $m = 11$ .
$y - 5 = 11(x - 3)$
Simplify the equation to the standard form $y = mx + c$ .
$y - 5 = 11x - 33$ $y = 11x - 33 + 5$ $y = 11x - 28$
The equation of the tangent to the curve at the point $(3, 5)$ is $\boxed{y = 11x - 28}$ .

Answer