Question

The line y=kx-5 is a tangent to the curve of f. Find the values of k

Answer 1

Answer:

k     =(5+b)/a

Step-by-step explanation:

Dear Student,

We know the line, y= kx-5

Curve is y = f(x)

Slope is given as,$\frac{d}{dx}f(x)$

$y = f^{'} (x)$

equation of tangent touches to the curve at (a,b)

(y-b) = dy/dx (x-a)

y = dy/dx(x-a) + b

kx-5 = dy/dx(x-a)+b

Let, m = dy/dx at (a,b)

kx-5 = m(x-a)+b

kx-5 = mx-ma+b

kx-5 = mx-(ma-b)

m = k

ma-b = 5

ka-b = 5

k     =(5+b)/a

Hope it helps

With Regards