Question

The tangents to the curve y = x2–5x+6 at the points (2,0) and (3,0) is -​

Answer 1

EXPLANATION.

Tangents to the curve,

⇒ y = x² - 5x + 6.

At the point = (2,0) & (3,0).

As we know that,

Slope of tangents of the curve.

⇒ dy/dx = d(x² - 5x + 6)/dx.

⇒ dy/dx = 2x - 5.

Slope at a point = (2,0).

⇒ m₁ = 2(2) - 5.

⇒ m₁ = 4 - 5.

⇒ m₁ = -1.

Slope at a point = (3,0).

⇒ m₂ = 2(3) - 5.

⇒ m₂ = 6 - 5.

⇒ m₂ = 1.

As we know  that,

⇒ m₁ x m₂ = -1.

⇒ (-1) x (1) = -1.

⇒ (-1) = (-1).

Tangents at the point (2,0) and (3,0) are at right angles.

                                                                                                                         

MORE INFORMATION.

Equation of tangent.

Equation of tangent to the curve y = f(x) at P(x₁, y₁) is,

(y - y₁) = m(x - x₁).

(1) = The tangent at (x₁, y₁) is parallel to x-axes = (dy/dx) = 0.

(2) = The tangent at (x₁, y₁) is parallel to y-axes = (dy/dx) = ∞.

(3) = The tangent line makes equal angles with the axes = (dy/dx) at (x₁, y₁) = ± 1.

Answer 2

Step-by-step explanation:

$\bf \color{blue} \: The \: equation \: of \: the \: curve \: is$

$\bf \color{blue} \: y = {x}^{2} - 5x + 6$

$\bf \color{blue} ∴ \frac{ \: dy \: }{ \: dx \: } = 2x - 5$

$\bf \color{blue} \: Let \: m_{1}, \: m_{2} \: be \: slopes \: of \: tangents \: to$

$\bf \color{blue} \: the \: curve \: at \: the \: points \:(2,0) \: (3,0).$

$\bf \color{blue} ∴ \: m_{1} = value \: of \: \frac{dy}{dx} \: at \: (2,0)$

$\bf \color{blue} \: \: \: \: \: \: \: \: \: \: \: \: \: = 2(2) - 5 = 4 - 5 = - 1$

$\bf \color{blue} ∴ \: m_{2} = value \: of \: \frac{dy}{dx} \: at \: (3,0)$

$\bf \color{blue} \: \: \: \: \: \: \: \: \: \: \: \: \: = 2(3) - 5 = 6 - 5 = 1$

$\bf \pink {=> \: m _{1}m_{2} = ( - 1)(1) = - 1}$

$\bf \color{blue} \: tangents \: to \: the \: given \: curve \: at \: (2,0) \:$

$\bf \color{blue} \:and \: (3,0) \: are \: perpendicular \: to \: each \: other.$