Question

The point at which the tangent to the curve y= √4x-3 -1 has its slope 2/3. urgent pls​

Answer 1

Step-by-step explanation:

Given that: The point at which the tangent to the curve y= √4x-3 -1 has its slope 2/3.

To find : Point on curve

Solution: We know that slope of tangent on curve is given by

dy/dx

here

$y = \sqrt{4x - 3 } - 1$

Differentiation it with respect to x

$\frac{dy}{dx} = \frac{4}{2 \sqrt{4x - 3} } - 0 \\ \\ \frac{dy}{dx} = \frac{2}{ \sqrt{4x - 3} }$

As slope of tangent is 2/3,so equate both the slopes

$\frac{2}{\sqrt{4x - 3} } = \frac{2}{3} \\ \\ \frac{1}{ \sqrt{4x - 3} } = \frac{1}{3} \\ \\ squaring \: both \: sides \\ \\ \frac{1}{4x - 3} = \frac{1}{9} \\ \\ 4x - 3 = 9 \\ \\ 4x = 12 \\ \\ x = \frac{12}{4} \\ \\ x = 3$

Now put x= 3 in the curve,to find the value of y

$y = \sqrt{4(3) - 3 } - 1 \\ \\ y =\sqrt{12 - 3} - 1 \\ \\ y =\sqrt{9} - 1 \\ \\ y =±3 - 1 \\ \\ y = 2 \\ \\ and \\y=-4$

Thus, the points are (3,2) and (3,-4)

Hope it helps you.

Answer 2

Answer:

the points are (3 ,2) and (3,-4)