Differentiate the given curve y = x3 + x2 + 1 to find the slope function.

Question

ScanSolve AI Tutor · Accepted Answer

Here's how to find the equation of the tangent for part (a):

Differentiate the given curve $y = x^3 + x^2 + 1$ to find the slope function.
$\frac{dy}{dx} = \frac{d}{dx}(x^3) + \frac{d}{dx}(x^2) + \frac{d}{dx}(1)$ $\frac{dy}{dx} = 3x^2 + 2x + 0$ $\frac{dy}{dx} = 3x^2 + 2x$
Evaluate the derivative at the given point $(1, 3)$ to find the slope of the tangent, $m_t$ .
Substitute $x=1$ into $\frac{dy}{dx}$ :
$m_t = 3(1)^2 + 2(1)$ $m_t = 3(1) + 2$ $m_t = 3 + 2$ $m_t = 5$
Use the point-slope form of a linear equation, $y - y_1 = m_t(x - x_1)$ , with the point $(x_1, y_1) = (1, 3)$ and the slope $m_t = 5$ .
$y - 3 = 5(x - 1)$
Simplify the equation to the slope-intercept form $y = mx + c$ .
$y - 3 = 5x - 5$ $y = 5x - 5 + 3$ $y = 5x - 2$
The equation of the tangent to the curve $y = x^3 + x^2 + 1$ at the point $(1, 3)$ is:

Answer