Answer:
See below.
Step-by-step explanation:
5x^2 + 55x + 140 ---> required
Coefficients a=5, b=55, c=140 yield
[tex]h=\frac{-h}{2*a} =\frac{-(55)}{2*5} =\frac{-55}{10} =-5.5[/tex]
Thus, the x-coordinate of the vertex is [tex]\boxed{h = -5.5 }[/tex]
Plugging x=-5.5 into the given equation yields:
[tex]k=5*(-5.5)^{2} +55*-5.5+140=-11.25[/tex]
Thus, the y-coordinate of the vertex is [tex]\boxed{k = -11.25 }[/tex]
Altogether, the vertex of the given parabola is
[tex]\boxed{ (h,k)=(-5.5,-11.25) }[/tex]
Plugging (h,k)=(−5.5,−11.25) into the vertex formula
[tex]\boxed{ y=(x-h)^2+k }[/tex]
yields the vertex form of the parabola
[tex]\boxed{ y=(x+5.5)^2-11.25 }[/tex]
- To find the vertex of a parabola in standard form, first, convert it to the vertex form y=a(x−h)2+k y = a ( x − h ) 2 + k .
