Question

The point on the curve y = 7x- 3x^2 where the inclination of the tangent is 45 degree ​

Answer 1

Answer:

The point at which the inclination of tangent of curve is 45° is (1,4)

Step-by-step explanation:

Given that,the curve is

$y = 7x - 3x {}^{2}$

To find inclination,we need to find the slope of tangent.The slope of tangent is given as

$m = \frac{dy}{dx} = \frac{d}{dx} (7x - 3x {}^{2} ) = 7 - 6x$

Given that,the inclination of the tangent is 45°.Therefore

$m = tan45 {}^{0} = 1 \\ = > 7 - 6x = 1 \\ x = 1$

y can also be found as

$y = 7x - 3x {}^{2} = 7(1) - 3(1) {}^{2} = 4$

Therefore the point at which the inclination of tangent of curve is 45° is (1,4)

#SPJ3

Answer 2

Answer:

Point = (1,4)

Step-by-step explanation:

y = 7x- 3x^2

dy/dx = -6x + 7

dy/dx = slope at any point x = tan(angel of slope  )

inclination of the tangent  = 45 °

-6x + 7 = tan(45°) = 1

6x = 6

x =1

when x = 1 ,

y = 7(1) - 3(1)^2

y = 4

point = (1,4)

#SPJ3