The angle between the tangent of the curve y=x2-5x+6at the point (2,0)and (3,0)?

Question

Answer

Angle between Tangents at (2,0) and (3,0) of the Curve y=x^2-5x+6

Let's first find the equation of the curve:

Equation of the Curve

The given curve is:

$$
y=x^2-5x+6
$$

To find the slope of the tangent at any point on the curve, we need to differentiate the equation of the curve with respect to x:

$$
\frac{dy}{dx} = 2x-5
$$

Slope of Tangent at (2,0)

At point (2,0), the slope of the tangent is:

$$
\frac{dy}{dx}\bigg|_{(2,0)} = 2(2)-5 = -1
$$

Slope of Tangent at (3,0)

At point (3,0), the slope of the tangent is:

$$
\frac{dy}{dx}\bigg|_{(3,0)} = 2(3)-5 = 1
$$

Angle between the Tangents

The angle between two lines with slopes m1 and m2 is given by:

$$
heta = an^{-1}(\frac{m2-m1}{1+m1m2})
$$

Using this formula, we can find the angle between the tangents at (2,0) and (3,0):

$$
heta = an^{-1}(\frac{1-(-1)}{1+(-1)(1)}) = an^{-1}(1) = \frac{\pi}{4} \approx 45^{\circ}
$$

Conclusion

Therefore, the angle between the tangents at (2,0) and (3,0) of the curve y=x^2-5x+6 is approximately 45 degrees.

Answer

At(2,0)slope m1=-1;at (3,0)slope m2=1;;;m1×m2=-1.....angle=90degrees

Answer

Yes bro

Answer

1+m1×m2=1-1=0......90 degrees