Respuesta :
Given:
The function are:
[tex]f(x)=Kx^2+4x-3[/tex]
[tex]g(x)=2x-7[/tex]
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,
[tex]f(x)=Kx^2+4x-3[/tex]
Differentiate this function with respect to x.
[tex]f'(x)=K(2x)+4(1)-(0)[/tex]
[tex]f'(x)=2Kx+4[/tex]
Let the point of tangency is [tex](x_0,y_0)[/tex]. So, the slope of the tangent line is:
[tex][f'(x)]_{(x_0,y_0)}=2Kx_0+4[/tex]
On comparing [tex]g(x)=2x-7[/tex] with slope-intercept form, we get
[tex]m=2[/tex]
So, the slope of the tangent line is 2.
[tex]2Kx_0+4=2[/tex]
[tex]2Kx_0=2-4[/tex]
[tex]x_0=\dfrac{-2}{2K}[/tex]
[tex]x_0=-\dfrac{1}{K}[/tex]
Putting [tex]x=x_0,g(x)=y_0[/tex] in g(x), we get
[tex]y_0=2x_0-7[/tex]
Putting [tex]x=-\dfrac{1}{K}[/tex] in the above equation, we get
[tex]y_0=2(-\dfrac{1}{K})-7[/tex]
[tex]y_0=-\dfrac{2}{K}-7[/tex]
Putting [tex]x=-\dfrac{1}{K}[/tex] and [tex]f(x)=-\dfrac{2}{K}-7[/tex] in f(x).
[tex]-\dfrac{2}{K}-7=K\left(-\dfrac{1}{K}\right)^2+4(-\dfrac{1}{K})-3[/tex]
[tex]-\dfrac{2}{K}-7=\dfrac{1}{K}-\dfrac{4}{K}-3[/tex]
[tex]-\dfrac{2}{K}-7=\dfrac{-3}{K}-3[/tex]
Multiply both sides by K.
[tex]-2-7K=-3-3K[/tex]
[tex]-2+3=7K-3k[/tex]
[tex]1=4k[/tex]
[tex]\dfrac{1}{4}=K[/tex]
Therefore, the value of K is [tex]\dfrac{1}{4}[/tex].
