Question 724015
1. {{{f(x)= -x^2+8x+7}}}...here you have parabola with {{{a=-1}}}, so parabola is facing  {{{downward}}} and the maximum {{{y-value}}} on the graph will be {{{y}}} coordinate of the vertex

 the equation is {{{y = (x - h)^2 + k}}} where {{{h}}} and {{{k}}} are {{{x}}} and {{{y}}} coordinate of the vertex

so, we need {{{k}}}

{{{f(x)= -x^2+8x+7}}}...factor completely

{{{f(x)= -x^2+8x+7}}}...replace {{{7}}} with {{{-16+23}}}

{{{f(x)=-x^2+8x-16+23}}}...group first three terms together

{{{f(x)=-(x^2-8x+16)+23}}}.......note that {{{-(x^2-8x+16)=-(x-4)^2}}}

{{{f(x)=-(x-4)^2+23}}}...vertex form...{{{h=4}}} and {{{k=23}}}

so, vertex is at {{{4,23}}} and the maximum {{{y-value}}} on the graph is {{{y=23}}}

see it on a graph:

{{{ drawing( 600,600, -5, 10, -10, 35,circle(4,23,0.1),locate(4,23-.2,"V(4,23)"),graph( 600,600, -5, 10, -10, 35, -x^2+8x+7)) }}}


2.

Using quadratic formula to find any {{{x-intercepts}}} on graph of equation.  {{{ y=x^2+6x-1}}}


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} ...note that {{{a=1}}}, {{{b=6}}}, and {{{c=-1}}}


{{{x = (-6 +- sqrt( 6^2-4*1*(-1) ))/(2*1) }}} 

{{{x = (-6 +- sqrt( 36+4 ))/2 }}}

{{{x = (-6 +- sqrt( 40 ))/2 }}}

{{{x = (-6 +- 6.3)/2 }}}

solutions:

{{{x = (-6 + 6.3)/2 }}}

{{{x = 0.3/2 }}}

{{{x = 0.15 }}}

and

{{{x = (-6 -6.3)/2 }}}

{{{x = -12.3/2 }}}

{{{x = -6.15 }}}

so, {{{x-intercepts}}} are at ({{{0.15}}},{{{0}}}) and ({{{-6.15}}},{{{0}}})

check it on a graph:


{{{ drawing( 600,600, -10, 5, -10, 10,circle(0.15,0,0.1),circle(-6.15,0,0.1),graph( 600,600, -10, 5, -10, 10, x^2+6x-1)) }}}



3. complete the square

{{{x^2+16x=-4}}}

{{{x^2+16x+4=0}}}...replace {{{4}}} with {{{64-60}}}

{{{x^2+16x+64-60=0}}}...group first three terms together

{{{(x^2+16x+64)-60=0}}}....note that {{{(x^2+16x+64)}}} is square of a sum

{{{(x+8)^2-60=0}}}