Question

7. Find the slop of the tangent to the curve.
y = x2-3x+2 at the point whose x-coordinate is 3.​

Answer 1

EXPLANATION.

$\sf : \implies \: slope \: of \: the \: tangent \: to \: the \: curve = y = {x}^{2} - 3x + 2 \\ \\ \sf : \implies \: at \: the \: point \: whose \: x - coordinate \: = 3$

$\sf : \implies \: as \: we \: know \: that \: \\ \\ \sf : \implies \: slope \: \: of \: tangent \: = \frac{dy}{dx} \\ \\ \sf : \implies \: \frac{dy}{dx} = 2x - 3 \\ \\ \sf : \implies \frac{dy}{dx} = 2(3) - 3 \\ \\ \sf : \implies \: \frac{dy}{dx} = 3 \\ \\ \sf : \implies \: slope \: of \: tangent = 3$

Also we can calculate the equation

of tangent.

$\sf : \implies \: \dfrac{dy}{dx} = 2x - 3 = 3 \\ \\ \sf : \implies \: 2x = 6 \\ \\ \sf : \implies \: x \: = 3 \: \\ \\ \sf : \implies \: put \: the \: value \: of \: x = 3 \: in \: equation \: \\ \\ \sf : \implies \: y = (3) {}^{2} - 3(3) + 2 \\ \\ \sf : \implies \: y = 2$

$\sf : \implies \: \: its \: coordinate \: = (3,2) \\ \\ \sf : \implies \: equation \: of \: tangent \\ \\ \sf : \implies \: (y - y_{1}) = \frac{dy}{dx}(x - x_{1}) \\ \\ \sf : \implies \: (y - 2) = 3(x - 3) \\ \\ \sf : \implies \: y - 2 = 3x - 9 \\ \\ \sf : \implies \: 3x - y = 7 = equation \: of \: tangent$