Question 40036: for quadratic functions, sketch two parabolas showing all procedure (concavity, y-intercept, x-intercept, symmetric axis and vertex point)
a. y=2x^2-3x+1
b.y=-3x^2+2x+1
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! For quadratic functions, sketch two parabolas showing
> all procedure(concavity, y-intercept, x-intercepts,
> symmetric axis, and vertex point)
> y=2x^2-3x+1...make x coefficient 1 and .complete square
y=2{x^2-2*x*0.75+0.75^2}+1-2*0.75^2
=2(x-0.75)^2-0.125...so this will have a minimum at x=0.75...then y =-0.125...so
1.vertex is(0.75,-0.125)...it is a minimum...so parabola has a trough
at the bottom or convex down wards .....concave upwards.
2.axis of symmetry is x-0.75=0....or x=0.75
3.when x =0....y=2*0.75^2-0.125=1...hence y intercept=1
4.when y=0....2(x-0.75)^2=0.125......so x is
REAL...that is it has 2 x interceptS..that is it cutS x
axis AT 2 PLACES
5.as far as sketching goes , you can try that by
1.draw x & y axes with a suitable scale.
2.plot vertex at (0.75,1.125)
3.draw axis of symmetry a vertical line at x=0.75
4.plot th y intercept at x=0 and y=2.25..that is curve cuts y axis at 2.25
5.sketch a symmetric graph from vertex upwards around the axis of
symmetry of x=0.75
> y=-3x^2+2x+1...same way as abov
y=-3{x^2-2*x*(1/3)+(1/3)^2}+1+(1/3)^2
=(10/9)-3{x-(1/3)}^2
so....
1.vertex is(1/3,10/9)...it is a maximum...so parabola has a peak at
the top or convex up wards .....concave downwards.
2.axis of symmetry is x-1/3=0....or x=1/3
3.when x =0....y=10/9-3/9=7/9...hence y intercept=7/9
4.when y=0....3{x-(1/3)}^2=10/9........so x is real=(1/3)+ and -
sqrt(10/27)...that is it has 2.. x intercepts..that is it cuts x axis
at 2 places
5.as far as sketching goes ,but you can try that by the same way as given above.
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