Question

1
At which point on the curve y = 3x - x? the slope is -5.
O (4,-4)
O (-4,4)
O (2,-2)
O (-2,2)
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Answer 1

QUESTION :-

At which point on the curve y = 3x - x² the slope is -5.

O (4,-4)

O (-4,4)

O (2,-2)

O (-2,2)

___________________

Solution :

$\implies \rm y = 3x - {x}^{2}$

Now , we will find dy/dx because dy/dx is slope of tangent to the curve.

$\implies \rm \dfrac{dy}{dx} = \dfrac{d(3x - {x}^{2} )}{dx}$

$\implies \rm \dfrac{dy}{dx} = \dfrac{d(3x )}{dx} - \dfrac{d( {x}^{2} )}{dx}$

$\implies \rm \dfrac{dy}{dx} =3 - 2x$

Now slope is given = -5

Let the point at which slope be -5 be (x_1 , y_1)

So, put dy/dx = -5

$\implies \rm - 5 =3 - 2x_ 1$

$\implies \rm - 5 - 3 = - 2x_ 1$

$\implies \rm - 8= - 2x_ 1$

$\implies \rm \dfrac{ - 8}{ - 2} = x_ 1$

$\implies \rm x_ 1 = 4$

Now , to find y_1 put x_1= 4 in y = 3x - x².

$\implies \rm y_1 = 3x_1 - {x_1}^{2}$

$\implies \rm y_1 = (3 \times 4) - {(4)}^{2}$

$\implies \rm y_1 =12 - 16$

$\implies \rm y_1 = - 4$

Point on the curve y = 3x - x² the slope is -5 = (4,-4)

Answer :

• first option (4,-4) is correct