I am working on Calculus I by Marsden and Weinstein and this is Exercise 43 from Section 1.6
Let $g(x)=-4x^2+8x+13$. Show that the linear approximation to $g(3+\Delta x)$ always gives an answer which is too large, regardless of whether $\Delta x$ is positive or negative. Interpret your answer geometrically by drawing a graph of $g$ and its tangent line when $x_0=3$.
So the linear approximation around 3 is given by $g(3+\Delta x)\approx 1-16\Delta x$ and the tangent line at $x=3$ is given by $L(x)=1-16(x-3)$. So we know that $g(3)=1$. So my question is, why does it say in the exercise that the approximation gives an answer that is too large regardless of the value of $\Delta x$. For example, I was playing around and for $\Delta x=0.1$, the approximation is -0.6, which one could say is somewhat off. For $\Delta x=-0.1$ the approximation is 2.6 which is way to off, but at least I could not notice the approximation being too large for smaller and smaller values of $\Delta x$. I did graph the function $g(x)$ along with the tangent line, but still cannot see anything. I would appreciate any hints.