# Thread: Equation of the Tangent

1. ## Equation of the Tangent

Hello! I have this question about tangents, which I have solved partly. But I have some difficulties in solving the ii - part. I would appreciate your help

Here is the question:
Consider the function h(x) = x^(1/5)
i) Find the equation of the tangent to the graph at the point where x = a, (a doesn't equal 0). Write the equation in the form y = mx + c.

I have solved this part. My approach was first to find the derivative of the function and then replace the x by a. And then convert it into the form
y = mx + c, which gave me this result:
y = 1/5 a^(-4/5) a + 4/5 a^(1/5)

The part I couldn't solve:
ii) Show that this tangent intersects the x-axis at the point (-4a, 0).
I would really appreciate your help.
Thanks,
Amine

Hello! I have this question about tangents, which I have solved partly. But I have some difficulties in solving the ii - part. I would appreciate your help

Here is the question:
Consider the function h(x) = x^(1/5)
i) Find the equation of the tangent to the graph at the point where x = a, (a doesn't equal 0). Write the equation in the form y = mx + c.

I have solved this part. My approach was first to find the derivative of the function and then replace the x by a. And then convert it into the form
y = mx + c, which gave me this result:
y = 1/5 a^(-4/5) a + 4/5 a^(1/5)

The part I couldn't solve:
ii) Show that this tangent intersects the x-axis at the point (-4a, 0).
I would really appreciate your help.
Thanks,
Amine

you have a little confusion: the tangency point is (a, a^(1/5)) and since the slope of the tangent line at this point is (1/5)a^(-4/5), then the tangent line's equation is:

y - a^(1/5) = (1/5)a^(-4/5)[x - a] ==> y = (1/5)a^(-4/5)*x +(4/5)a^(1/5)

It's almost the same you got, but for some reaosn you put a where x should be!

Now just put y = 0 in the equation aboe and find out what the value of x is.

Tonio

3. Originally Posted by tonio
you have a little confusion: the tangency point is (a, a^(1/5)) and since the slope of the tangent line at this point is (1/5)a^(-4/5), then the tangent line's equation is:

y - a^(1/5) = (1/5)a^(-4/5)[x - a] ==> y = (1/5)a^(-4/5)*x +(4/5)a^(1/5)

It's almost the same you got, but for some reaosn you put a where x should be!

Now just put y = 0 in the equation aboe and find out what the value of x is.

Tonio
Hey,
so should I say 0 = (1/5)a^(-4/5)*x +(4/5)a^(1/5)
and then solve for x?
Thanks a lot for your help