## University Calculus: Early Transcendentals (3rd Edition)

The parabola does have a tangent line whose slope is $-1$. The point of tangency is $(3,-16)$. And the equation of the tangent line is $y=-x-13$.
$$y=f(x)=2x^2-13x+5$$ First, we will find the derivative $y'$: $$y'=\lim_{z\to x}\frac{f(z)-f(x)}{z-x}=\lim_{z\to x}\frac{(2z^2-13z+5)-(2x^2-13x+5)}{z-x}$$ $$y'=\lim_{z\to x}\frac{(2z^2-2x^2)+(-13z+13x)}{z-x}=\lim_{z\to x}\frac{2(z-x)(z+x)-13(z-x)}{z-x}$$ $$y'=\lim_{z\to x}\Big(2(z+x)-13\Big)$$ $$y'=2(x+x)-13=4x-13$$ $y'(x_0)$ is also the slope of tangent line to $f(x)$ at $x=x_0$. Therefore, to have a tangent line with slope $-1$ means $$4x_0-13=-1$$ $$x_0=3$$ For $x_0=3$, $f(3)=2\times3^2-13\times3+5=18-39+5=-16$ So the parabola does have a tangent line whose slope is $-1$. The point of tangency is $(3,-16)$. The equation of the tangent line whose slope is $-1$ will have this form: $$y=-x+m$$ To find $m$, we substitute the coordinates of $(3,-16)$ here: $$-3+m=-16$$ $$m=-13$$ Therefore, the equation of the tangent line is $$y=-x-13$$