Question

f(x)=(x-2)^1/2
f'(x)=1/2(x-2)^1/2
Find equations of the tangent line(s) that pass through the point (5,2). (the point (5,2) is NOT on the graph of f.)

Answer 1

Definition of the derivative was used to get f'(x). Which is the slope.

Answer 2

f(x)=(x-2)^1/2
so f'(x)=1/2*(x-2)^(-1/2)
we have to find equation of a tangent which passes through 5,2
let the equation be y=mx+c
2=5m+c
now the tangent is touching f(x) at a,b
so b=(a-2)^1/2
and the slope at this point is the slope of the tangent
so slope m=1/2(a-2)^(-1/2)
also (a,b) satisfy the tangent equation
so b=am+c
now we have 4 equations and 4 variables a, b, m and c
eq.1 b=(a-2)^1/2
eq.2 b=am+c
eq.3 m=1/2(a-2)^(-1/2)
eq 4. 2=5m+c
substitute 1 in 2, then 4 in 2 and then 3 in 2
we get
\[\sqrt(a-2)=a/2*1/\sqrt(a-2)+2-5/2\sqrt(a-2)\]
now put a-2=z^2
we'll get a quadratic equation in z
z^2-4z+3=0
solving we get z=1,3
I proceeded with z=1
a-2=z^2
so a=3
so m=1/2*(a-2)^(-1/2)
so slope =1/2
we have 2=5m+c
putting value of m we get c=-1/2
so equation of tangent is
y=m/2-1/2

Answer 3

let m := f'(5), which is 0.5 * sqrt(3)
\[y - y_0 = m(x - x_0)\]plug in 2 for y_0 and 5 for x_0
\[y = \frac{\sqrt{3}}{2}x - \frac{5\sqrt{3}}{2} + \frac{4}{2}\]

Answer 4

see the graph

Answer 5

you can see from graph the point where tangent touches the curve f(x) slope is 1/2

Answer 6

Thank you very much! This was really helpful!

Answer 7

welcome, glad to help u