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Question 537951: What is the graph of this functions?
y= 1/2x^2 -3x+11/2
y= -2x^2+4x+3
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! What is the graph of this functions?
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Problem 1.
y= 1/2x^2 -3x+11/2
complete the square
y=(1/2)(x^2-6x+9)+11/2-9/2
y=(1/2)(x-3)^2+1
This is an equation of a parabola of standard form: y=A(x-h)^2+k, with (h,k) being the (x,y) coordinates of the vertex. Parabola opens upwards.
For given equation:
vertex: (3,1)
A=1/2
x-intercepts
set y=0
(1/2)(x-3)^2+1=0
(x-3)^2=-2
x-intercepts: none
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y-intercept
set x=0
y=1/2x^2 -3x+11/2
y=11/2
Because the parabola is symmetrical about its axis of symmetry(x=3), this gives another point (6,11/2)
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With coordinates of the vertex, the y-intercept, point (6,11/2), no x-intercepts and the knowledge the parabola opens upwards, you should easily be able to graph given function
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Problem 2.
y= -2x^2+4x+3
complete the square
y=-2(x^2-2x+1)+3+2
y=-2(x-1)^2+5
This is an equation of a parabola of standard form: y=-A(x-h)^2+k, with (h,k) being the (x,y) coordinates of the vertex. Parabola opens downwards.
For given equation:
vertex: (1,5)
A=-2
x-intercepts
set y=0
-2(x-1)^2+5=0
-2(x-1)^2=-5
(x-1)^2=5/2
take sqrt of both sides
x-1=±√(5/2)≈±1.58
x=1±1.58
x=2.58 and -0.58
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y-intercept
set x=0
y=-2x^2+4x+3=3
Because the parabola is symmetrical about its axis of symmetry(x=1), this gives another point (2,3)
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With coordinates of the vertex, the y-intercept, point (2,3), x-intercepts, and the knowledge the parabola opens downwards, you should easily be able to graph given function.
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See the following graphs as visual checks on the answers:
Because the parabola is symmetrical about its axis of symmetry(x=3), this gives another point (6,11/2)
With coordinates of the vertex, the y-intercept, point (6,11/2) and the knowledge the parabola opens upwards, you should easily be able to graph given function. See graph below as a visual check on the answers. Red curve for problem 1 and green curve for problem 2
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