Find the largest real number $c$ such that $1$ is in the range of $f(x)=x^2-5x+c+x^2-8x$.

Answer 1

We have $f(x)=2x^2-13x+c$. The range of $f$ is $\{y \in \mathbb{R}\mid y \ge d\}$, where $d$ is the discriminant of the quadratic.

Because the quadratic has a positive leading coefficient, the range is $[d,\infty).$ The discriminant is $(-13)^2-4(2)(c)=169-8c$, and for $d \ge 1$ we must have $169-8c \le 0.$ Solving gives $c \ge 21.125,$ so the largest such $c$ is $\boxed{21.125}$.