Question 180892


In order to find the vertex, we first need to find the x-coordinate of the vertex.



To find the x-coordinate of the vertex, use this formula: {{{x=(-b)/(2a)}}}.



{{{x=(-b)/(2a)}}} Start with the given formula.



From {{{y=x^2+6x+5}}}, we can see that {{{a=1}}}, {{{b=6}}}, and {{{c=5}}}.



{{{x=(-(6))/(2(1))}}} Plug in {{{a=1}}} and {{{b=6}}}.



{{{x=(-6)/(2)}}} Multiply 2 and {{{1}}} to get {{{2}}}.



{{{x=-3}}} Divide.



So the x-coordinate of the vertex is {{{x=-3}}}. Note: this means that the axis of symmetry is also {{{x=-3}}}.



Now that we know the x-coordinate of the vertex, we can use it to find the y-coordinate of the vertex.



{{{f(-3)=x^2+6x+5}}} Start with the given equation.



{{{f(-3)=(-3)^2+6(-3)+5}}} Plug in {{{x=-3}}}.



{{{f(-3)=1(9)+6(-3)+5}}} Square {{{-3}}} to get {{{9}}}.



{{{f(-3)=9+6(-3)+5}}} Multiply {{{1}}} and {{{9}}} to get {{{9}}}.



{{{f(-3)=9-18+5}}} Multiply {{{6}}} and {{{-3}}} to get {{{-18}}}.



{{{f(-3)=-4}}} Combine like terms.



So the y-coordinate of the vertex is {{{y=-4}}}.



So the vertex is *[Tex \LARGE \left(-3,-4\right)].



Now because {{{a=1}}} (which is positive), this means that the parabola opens upward and that there is a minimum.


So the min is the same as the y-coordinate of the vertex. This means that the min is {{{y=-4}}}



Here's a graph to confirm our answers:



{{{ graph( 500, 500, -10, 10, -10, 10, x^2+6x+5) }}} Graph of {{{y=x^2+6x+5}}}