Question 111879
First let's find the vertex. To find the vertex, we first need the axis of symmetry (ie the x-coordinate of the vertex)



To find the axis of symmetry, use this formula:


{{{x=-b/(2a)}}}


From the equation {{{y=-3x^2+x-5}}} we can see that a=-3 and b=1


{{{x=(-1)/(2*-3)}}} Plug in b=1 and a=-3



{{{x=(-1)/-6}}} Multiply 2 and -3 to get -6




{{{x=1/6}}} Reduce



So the axis of symmetry is  {{{x=1/6}}}



So the x-coordinate of the vertex is {{{x=1/6}}}. Lets plug this into the equation to find the y-coordinate of the vertex.


Lets evaluate {{{f(1/6)}}}


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



{{{f(1/6)=-3(1/6)^2+(1/6)-5}}} Plug in {{{x=1/6}}}



{{{f(1/6)=-3(1/36)+(1/6)-5}}} Square {{{1/6}}} to get {{{1/36}}}



{{{f(1/6)=-3/36+(1/6)-5}}} Multiply




{{{f(1/6)=-1/12+1/6-5}}} Reduce



{{{f(1/6)=-59/12}}} Combine like terms



So the y-coordinate of the vertex is {{{y=-59/12}}}


So the vertex is *[Tex \LARGE \left(\frac{1}{6},\frac{-59}{12}\right)]




Now let's find the x-intercepts. To find the x-intercepts, let y=0 and solve for x


{{{0=-3x^2+x-5}}} Set y equal to zero to solve for x



*[invoke quadratic_formula -3, 1, -5, "x"]