Question

Determine the value of K that will cause f(x)=Kx^2+4x-3 to intersect the line g(x)=2x-7 at one point. SHOW ALL YOUR STEPS, DON'T USE DECIMALS INSTEAD USE FRACTIONS PLEASE!!!!!

Answer 1

Given:

The function are:

`f(x)=Kx^2+4x-3`

`g(x)=2x-7`

The graph of f(x) intersect the line g(x) at one point.

To find:

The value of K.

Solution:

The graph of f(x) intersect the line g(x) at one point. It means the line g(x) is the tangent line.

We have,

`f(x)=Kx^2+4x-3`

Differentiate this function with respect to x.

`f'(x)=K(2x)+4(1)-(0)`

`f'(x)=2Kx+4`

Let the point of tangency is
`(x_0,y_0)`. So, the slope of the tangent line is:

`[f'(x)]_((x_0,y_0))=2Kx_0+4`

On comparing
`g(x)=2x-7` with slope-intercept form, we get

`m=2`

So, the slope of the tangent line is 2.

`2Kx_0+4=2`

`2Kx_0=2-4`

`x_0=(-2)/(2K)`

`x_0=-(1)/(K)`

Putting
`x=x_0,g(x)=y_0` in g(x), we get

`y_0=2x_0-7`

Putting
`x=-(1)/(K)` in the above equation, we get

`y_0=2(-(1)/(K))-7`

`y_0=-(2)/(K)-7`

Putting
`x=-(1)/(K)` and
`f(x)=-(2)/(K)-7` in f(x).

`-(2)/(K)-7=K\left(-(1)/(K)\right)^2+4(-(1)/(K))-3`

`-(2)/(K)-7=(1)/(K)-(4)/(K)-3`

`-(2)/(K)-7=(-3)/(K)-3`

Multiply both sides by K.

`-2-7K=-3-3K`

`-2+3=7K-3k`

`1=4k`

`(1)/(4)=K`

Therefore, the value of K is
`(1)/(4)`.