# Thread: Tangent to a curve

1. ## Tangent to a curve

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

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

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

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

2. ## Re: Tangent to a curve

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

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

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

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

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

$\displaystyle f'(x) = 2x$

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

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

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

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

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

3. ## Re: Tangent to a curve

An alternative solution:

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

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

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