Question 62196
Graph Y=3x^2-8x+2 and give the vertex, domain and range?

The y-intercept happens when x=0
y=3(0)^2-8(0)+2
y=0-0+2
y=2  Plot (0,2)
:
The quadratic equation is in standard form {{{y=ax^2+bx+c}}}
a=3, b=-8, and c=2
Because a is positive, the parabola opens upward.
The x coordinate of the vertex is found with the formula {{{x=-b/2a}}}
{{{x=-(-8)/(2(3))}}}
{{{x=8/6}}}
{{{x=4/3}}}
Substitute that into the equation to find the y coordinate.
{{{y=3(4/3)^2-8(4/3)+2}}}
{{{y=3(16/9)-32/3+2}}}
{{{y=16/3-32/3+2}}}
{{{y=16/3-32/3+6/3}}}
{{{y=-10/3}}}
The vertex is (4/3,-10/3)
The domain for all parabolas is all real numbers. {x|x=R}
The range is from the vertex upward since it opens upward. {y|y>=-10/3}
{{{graph(300,200,-10,10,-10,10,3x^2-8x+2)}}}
Happy Calculating!!!