Answer
Yes, $y=-x-13$ at $(3,-16)$.
Work Step by Step
Step 1. Given $y=f(x)=2x^2-13x+5$, we have
$y'=f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{2(x+h)^2-13(x+h)+5-2x^2+13x-5}{h}=\lim_{h\to0}\frac{4xh+2h^2-13h}{h}=\lim_{h\to0}(4x+2h-13)=4x-13$
Step 2. Let $y'=-1$; we have $4x-13=-1$, which gives $x=3$
Step 3. The point on the curve with $x=3$ is $y=2(3)^2-13(3)+5=-16$, or $(3,-16)$
Step 4. With the slope $m=-1$, the tangent line equation can be written as $y+16=-1(x-3)$ which gives $y=-x-13$ at the point of tangency of $(3,-16)$.
