### Math Notes

Subjects

#### Differential Calculus Solutions

##### Topics || Problems

Find the slope of the curve $$y = (x+1)(x+2)$$ at points where it crosses the x-axis. Trace the curve.

When the curve crosses the $$x-axis$$ the value of $$y = 0$$.

If $$y = 0$$ then $$0= x^2 +3x +2$$

$$x_1 = \frac{-3+\sqrt{9-4(2)}}{2} = -1$$

$$x_2 = \frac{-3-\sqrt{9-4(2)}}{2} = -2$$

$$y = (x+1)(x+2)$$

$$\frac{dy}{dx} = (x+1) +(x+2)$$

At $$x_1 = -1$$

$$\frac{dy}{dx} = 0+1 = 1$$

At $$x_1 = -2$$

$$\frac{dy}{dx} =-1+0 = -1$$

Thus the slope of the curve at $$(-1,0)$$ is 1 and -1 at $$(-2,0)$$