How do you find the axis of symmetry, and the maximum or minimum value of the function #v(t) = t^2 + 11t - 4#?

1 Answer
Feb 3, 2018

#x=-11/2,"minimum at "-137/4#

Explanation:

#"for a quadratic in standard form"#

#•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0#

#"then the x-coordinate of the vertex which is also the axis"#
#"of symmetry is"#

#•color(white)(x)x_(color(red)"vertex")=-b/(2a)#

#v(t)=t^2+11t-4" is in standard form"#

#"with "a=1,b=11" and "c=-4#

#rArrx_(color(red)"vertex")=-11/2#

#rArr"equation of axis of symmetry is "x=-11/2#

#"substitute "x=-11/2" into equation for y"#

#y_(color(red)"vertex")=(-11/2)^2+11(-11/2)-4=-137/4#

#rArrcolor(magenta)"vertex "=(-11/2,-137/4)#

#"the max/min value occurs at the vertex"#

#"for max/min consider the value of the coefficient a"#

#• " if "a>0" then minimum value"#

#• " if "a<0" then maximum value"#

#"here "a>0" hence minimum value"#

#"the minimum value "=-137/4#