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Suppose we have two polynomials of degree $n$ and $n-1$,
begin{align*}
f(x) &= a_n x^n + a_{n-1} x^{n-1} + dots + a_0, \
g(x) &= b_{n-1}x^{n-1} + b_{n-2} x^{n-2} + dots + b_0,
end{align*}
with $a_{n} neq 0$ and $b_{n-1} neq 0$. Let us consider the convex combination of the two polynomials $h(x,t) = (1-t) f(x) + t g(x)$. I read a statement that says: as $t to 1$, one of the zeros $h(x,t)$ would go to infinity. Intuitively, I think this is true because by Vieta's formula, as $t to 1$, the coefficient with $x^n$ would go to $0$ and it must be that one of the root should be unbounded. How do we argue this point?

It is a consequence of Rouche's theorem. We can prove that for all sufficiently large $R>0$, there exists $delta>0$ such that for each $t$ with $0<|t|<delta$, there exists unique $z=z(t)$ such that
$$
|z|geq R, ;tf(z) + (1-t)g(z) =0.
$$ Without loss of generality, let $$
R>max{|lambda_1|,| lambda_2|,ldots, |lambda_{n-1}|},$$
where $lambda_1, lambda_2,ldots, lambda_{n-1}$ are roots of $g(lambda)=0$ (counted with multiplicity.) And let $h_t(z)= g(z) +t(f(z)-g(z)).$ Now, consider
$$
N:tmapsto frac{1}{2pi i}int_{|z|=R} frac{h_t'(z)}{h_t(z)}dz.
$$ If $h_t$ does not vanish on $|z|=R$, then $N$ gives us the number of roots of $h_t=0$ in the region $|z|<R$ by Cauchy's argument principle. Note that if
$$
|t|<delta:=frac{min_{|z|=R} |g(z)|}{max_{|z|=R}|f(z)-g(z)|} leq min_{|z|=R}frac{|g(z)|}{|f(z)-g(z)|},
$$ then $h_t$ does not vanish on $|z|=R$, and $N:tin (-delta,delta)to mathbb{N}cup {0}$ defines a continuous integer-valued function. Since it holds that $N(0)= n-1$ by the choice of $R$, we have $$N equiv n-1$$ on $|t|<delta$. Note that if $tneq 0$, then by fundamental theorem of algebra, $h_t$ has $n$-roots counted with multiplicity. This implies that for all $0<|t|<delta$, there exists exactly one root $z_t in mathbb{C}setminus mathbb{D}(0,R)$ of $h_t(z)=0$ as desired.

Consider the reverse polynomials
begin{align}
tilde h(w,t)&=w^nh(w^{-1},t)=(1−t)x^nf(w^{-1})+tx^ng(w^{-1})\
&=
begin{array}{lll}
&(1−t)&(a_n+&a_{n-1}w+...+a_1w^{n-1}+a_0w^n)\
+&t~&(&b_{n-1}w+...+b_1w^{n-1}+b_0w^n)
end{array}
\
&=(1-t)tilde f(w)+twtilde g(w)
end{align}
Then one easily sees that $tilde h(w,t)$ has always $n$ roots $w$. These correspond to the roots of $h(t,x)$ via $x=w^{-1}$ for $t<1$. At time $t=1$ the polynomial $tilde h(w,1)=wtilde g(w)$ has $w=0$ and the inverses of the roots of $g$ as roots. The set of roots of $tilde h$ is a continuous function of $t$. As $tilde h$ has a continuous path $w_0(t)$, $tin [a,1]$ of roots converging to $0$, the original homotopy $h$ has a path $x_0(t)=w_0(t)^{-1}$, $tin [a,1)$, of roots that diverges to infinity.