Question 317033
Vertex form. Complete the square to get there.
{{{f(x)=x^2-7x-8}}}
{{{f(x)=x^2-7x+49/4-8-49/4}}}
{{{f(x)=(x-7/2)^2-32/4-49/4}}}
{{{f(x)=(x-7/2)^2-81/4}}}
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Vertex is ({{{7/2}}},{{{-81/4}}}).
Axis of symmetry is {{{x=7/2}}}
Coefficient of {{{x^2}}} term is positive, parabola opens upwards. 
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Solve to get the x-intercepts,
{{{f(x)=(x-7/2)^2-81/4=0}}}
{{{(x-7/2)^2=81/4}}}
{{{x-7/2=0 +- 9/2}}}
{{{x=7/2 +- 9/2}}}
{{{x=8}}} and {{{x=-1}}}
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{{{drawing(300,300,-2,10,-30,10,grid(1),
circle(7/2,-81/4,0.3),
circle(8,0,0.3),
circle(-1,0,0.3),
blue(line(7/2,-50,7/2,50)),
graph(300,300,-2,10,-30,10, x^2-7x-8))}}}