Question 640564: Give the following quadratic equation, find at the vertex, the axis of symmetry, the intercepts, the domain, the range, the interval where the function is increasing, the interval where the function is decreasing.
y=6x^2+10x+3
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Give the following quadratic equation, find at the vertex, the axis of symmetry, the intercepts, the domain, the range, the interval where the function is increasing, the interval where the function is decreasing.
y=6x^2+10x+3
complete the square
y=6(x^2+5x/3+25/36)+3-25/6
y=6(x^2+5/6)^2+18/6-25/6
y=6(x^2+5/6)^2-7/6
This is an equation of a parabola that opens upwards
Its standard form: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex.
vertex: (-5/6,-7/6)
axis of symmetry: x=-5/6
..
y-intercept
set x=0
y=0+0+3=3
..
x-intercepts
set y=0
6x^2+10x+3=0
solve for x by quadratic formula: 
a=6, b=10, c=3
x≈-1.274
or
x≈-.392
..
Range: [-7/6,∞)
Domain: (-∞,∞)
..
Interval function decreasing: (∞-5/6)
Interval function increasing: (-5/6,∞)
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