Question 214894
For:
{{{-x(x-3)}}} rewrite as {{{y = -x^2+3x}}}
Find:
a) The graph opens downward as indicated by the negative value of the {{{x^2}}} term.
b) The axis of symmetry (AOS) is given by:
{{{x = -b/2a}}} where: a = -1 and b = 3.
{{{x = (-3)/2(-1)}}}
{{{highlight(x = 3/2)}}} This is the equation of the AOS.
c) The vertex is lies on the AOS and is found by using the equation of the AOS for the x-coordinate, then substituting this into the given quadratic equation to solve for the corresponding y-coordinate.
{{{y = -x^2+3x}}} Substitute {{{x = 3/2}}}
{{{y = -(3/2)^2+3(3/2)}}} Evaluate.
{{{y = -(9/4)+(9/2)}}}
{{{y = 9/4}}}
The vertex is at: (3/2, 9/4)
d) 
The y-intercept is found by setting x = 0 in the given quadratic equation and solving for y.
{{{y = -x^2+3x}}} Substitute x = 0.
{{{y = 0}}}
The y-intercept is at: (0, 0)
e)
The zeros are found by setting the given quadratic equation equal to zero and solving.
{{{-x^2+3x = 0}}} Factor a -x.
{{{-x(x-3) = 0}}} Apply the zero product rule:
{{{-x = 0}}} or {{{x-3 = 0}}} so that...
{{{highlight(x = 0)}}} or {{{highlight(x = 3)}}}
The zeros are at (0, 0) and (3, 0)
f) 
Graph: {{{y = -x^2+3x}}}
{{{graph(400,400,-5,5,-5,5,-x^2+3x)}}}