Question 539861
{{{Y=3X^2-6X-5}}} intercept the y-axis (which is the {{{X=0}}} line)
at {{{Y=-5}}}. So the y-intercept is the point is (0,-5).
Because {{{3(X-1)^2-8=3(x^2-2x+1)-8=3X^2-6X+3-8=3X^2-6X-5}}},
The equation can be re-written as
{{{Y=3(X-1)^2-8}}},
showing that the vertex is the point with {{{X=1}}} and {{{Y=-8}}}.
So the vertex is the point (1,-8).
The x-intercepts are the points where the parabola crosses the x-axis (which is the {{{Y=0}}} line).
At those points, substituting {{{Y=0}}} in the modified equation, we see that
{{{0=3(X-1)^2-8}}} <--> {{{3(X-1)^2=8}}} <--> {{{(X-1)^2=8/3}}}
That gives us two solutions
{{{X-1=sqrt(8/3)}}} and {{{X-1=sqrt(8/3)}}}
or {{{X=1 +- sqrt(8/3)}}}
Multiplying and dividing by 6 we get a more popular way of expressing the solutions, which can be further simplified
{{{X=(6+- sqrt(96))/6=(6+- 4sqrt(6))/6=(3+- 2sqrt(6))/3}}}
The x-intercepts are the points with the above values for x and Y=0.