# Thread: Tangent to a curve

1. ## Tangent to a curve

Consider the function $f(x)= x^2+k, k\in \mathbb{R}$ and the tangent to the curve at $x=a$

Show that the tangent does not pass through the origin if $k<0$ .

It seems obvious that it does that it pass through the origin when I draw the function for $k<0$ as the origin is 'inside' the parabola. But is this argument enough?

I also have the equation of the tangent being $\displaystyle y = f'(a)x+c \implies y = 2ax+c$

2. ## Re: Tangent to a curve

Originally Posted by Bushy
Consider the function $f(x)= x^2+k, k\in \mathbb{R}$ and the tangent to the curve at $x=a$

Show that the tangent does not pass through the origin if $k<0$ .

It seems obvious that it does that it pass through the origin when I draw the function for $k<0$ as the origin is 'inside' the parabola. But is this argument enough?

I also have the equation of the tangent being $\displaystyle y = f'(a)x+c \implies y = 2ax+c$
you need to show the y-intercept of the tangent line cannot equal zero.

$f(x) = x^2+k \, ; \, k < 0$

$f'(x) = 2x$

at the point on the curve $(a, a^2+k)$ , the tangent line equation is

$y - (a^2+k) = 2a(x - a)$

$y = 2ax - 2a^2 + (a^2 + k)$

$y = 2ax - (a^2 - k)$

$k < 0 \implies (a^2 - k) \ne 0$

3. ## Re: Tangent to a curve

An alternative solution:

At x = a, the tangent line goes through the point $(a, a^2 + k)$ and the slope is $2a$. Therefore, writing the equation of the line in slope-intercept form,

$a^2 + k = 2a(a) + b$, where b is the y-intercept.

$k = a^2 + b$. If b = 0, then we get $k = a^2$, which cannot be true since k < 0 and the LHS is always non-negative.