Question 188556
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Given:{{{system(5x(2x-3))}}} --->  {{{10x^2-15x}}},--> where{{{system(a=10,b=(-15),c=0)}}}


Solving for the X-Intercepts:
Disciminant: {{{b^2-4ac=-15^2-4(10)(0)=225}}}
Then, by Quadratic,
{{{x=(-(-15)+-sqrt(225))/(2*10)=(15+-15)/20}}}
{{{x=(15+15)/20=30/20=red(1.5)}}}
{{{x=(15-15)/20=0/20=red(0)}}}


X-Intercepts, (1.5,0) & (0,0)


Let fx=0 ----> {{{y=10(0)^2-15(0)=red(0)}}}


Y-Intercept, (0,0)


For the vertex: {{{x=-b/2a=-(-15)/(2*10)=15/20=red(3/4)}}}
For Y, subst. x= 3/4 to the eqn:
{{{y=10(.75)^2-15(.75)=5.625-11.25=red(-5.625)}}}


The vertex, <font color=red>(3/4, -5.625)</font>

We'll see the graph:
{{{drawing(400,400,-3,5,-8,5,grid(1),graph(400,400,-3,5,-8,5),blue(circle(.75,-5.625,.08)),circle(1.5,0,.07),circle(0,0,.07),circle(0,0,.05),circle(1.5,0,.05),blue(circle(.75,-5.625,.06)))}}} ---> {{{drawing(400,400,-3,5,-8,5,grid(1),graph(400,400,-3,5,-8,5,10x^2-15x),blue(circle(.75,-5.625,.08)),circle(1.5,0,.07),circle(0,0,.07),circle(0,0,.05),circle(1.5,0,.05),blue(circle(.75,-5.625,.06)))}}}


Thank you,
Jojo</font>