[tex]\bf f(x) = -x^2-8x\implies f(x) = -x^2-8x+0 \\\\[-0.35em] ~\dotfill\\\\ \textit{vertex of a vertical parabola, using coefficients} \\\\ f(x)=\stackrel{\stackrel{a}{\downarrow }}{-1}x^2\stackrel{\stackrel{b}{\downarrow }}{-8}x\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{-8}{2(-1)}~~,~~0-\cfrac{(-8)^2}{4(-1)} \right)\implies \left( -4~~,~~0+\cfrac{64}{4} \right)\implies (-4~,~16)[/tex]
Answer:
Step-by-step explanation:
[tex]\text{The vertex form of a parabola}\ f(x)=ax^2+bx+c:\\\\y=a(x-h)^2+k\\\\(h,\ k)\ -\ \text{vertex}\\\\\bold{METHOD\ 1}\\\\\text{complete to the perfect square}\ (a+b)^2=a^2+2ab+b^2\qquad (*)\\\\f(x)=-x^2-8x=-x^2-2(x)(4)=-x^2-2(x)(4)-4^2+4^2\\\\=-(\underbrace{x^2+2(x)(4)+4^2}_{(*)})+4^2=-(x+4)^2+16=-(x-(-4))^2+16\\\\h=-4,\ k=16\to\text{vetrex}=(-4,\ 16)[/tex]
[tex]\bold{METHOD\ 2}\\\\\text{use the formula:}\\\\h=\dfrac{-b}{2a},\ k=f(h)\\\\f(x)=-x^2-8x\to a=-1,\ b=-8,\ c=0\\\\\text{substitute:}\\\\h=\dfrac{-(-8)}{2(-1)}=\dfrac{8}{-2}=-4\\\\k=f(-4)=-(-4)^2-8(-4)=-16+32=16\\\\\text{vertex}=(-4,\ 16)[/tex]
[tex]\bold{METHOD\ 3}\\\\\text{Find the zeros of}\ f(x):\\\\f(x)=-x^2-8x\\\\f(x)=0\to-x^2-8x=0\\\\-x(x+8)=0\iff -x=0\ \vee\ x+8=0\\\\x=0\ \vee\ x=-8\\\\h\ \text{is halfway between zeros}.\\\\h=\dfrac{0+(-8)}{2}=\dfrac{-8}{2}=-4\\\\k=f(h)\to k=f(-4)=-(-4)^2-8(-4)=-16+32=16\\\\\text{vertex}=(-4,\ 16)[/tex]
