Let $f(x)=x^2$ . Find the abscissa of the intersection point of the two tangent lines of $f(x)$ at $x=-4$ and at $x=2$.
I know I'm meant to find the two gradients of the two lines and use simultaneous equations to substitute the values, but I'm not sure how. Any help would be greatly appreciated.
Yes, your thoughts are right. We would start by finding the gradient of the two tangents by taking the derivative of $f(x) = x^2$, which gives $f'(x) = 2x$. From the definition of the derivative, $f'(x)$ is simply the gradient of the line tangent to $f(x)$ at a point with an abscissa of $x$. Thus, the gradient of $f(x)$ at $x = -4$ is $f'(-4) = 2 \cdot -4 = -8$, and the gradient at $x = 2$ is $f'(2) = 2 \cdot 2 = 4$. We can use point-slope form to get the equations of the tangents and find their intersection, but we first need to find the two points that the parabola intersects the tangents. Since we are given $f(x) = x^2$, we have $f(-4) = (-4)^2 = 16$ and $f(2) = 2^2 = 4$, meaning that the two points are $(-4, 16)$ and $(2, 4)$. Now, we use point-slope form to get that the equations of the two lines are $y - 16 = -8(x + 4)$ and $y - 4 = 4(x - 2)$. We now convert these equations into slope intercept form by expanding and isolating the $y$, which gives the equations $y = -8x - 16$ and $y = 4x - 4$. Solving gives: \begin{align*}-8x - 16 &= 4x - 4 \\ -12x &= 12 \\ x &= -1\end{align*}Since the question asked for the abscissa ($x$ coordinate), the answer should be $\boxed{1}$. We can also calculate the $y$ coordinate to be $y = 4(-1) - 4 = -4 - 4 = 8$. To confirm this answer, I also plotted $f(x)$ and its tangents on Desmos: $$ $$ Graph of f(x) and tangents $$ $$ Hope this helps!