we know that
The equation of a vertical parabola in vertex form is equal to
[tex]y=a(x-h)^{2}+k[/tex]
where
(h,k) is the vertex
if [tex]a>0[/tex] ----> the parabola open upward (vertex is a minimum)
if [tex]a<0[/tex] ----> the parabola open downward (vertex is a maximum)
case A) [tex]y=-x^{2}+5[/tex]
In this problem we have
the vertex is the point [tex](0,5)[/tex]
[tex]a=-1[/tex]
therefore
the parabola open downward (vertex is a maximum)
The graph with its corresponding equation in the attached figure
case B) [tex]y=\frac{3}{2} x^{2}-4x[/tex]
convert to vertex form
Factor the leading coefficient
[tex]y=\frac{3}{2}(x^{2}-\frac{8}{3}x)[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side.
[tex]y+\frac{8}{3}=\frac{3}{2}(x^{2}-\frac{8}{3}x+\frac{64}{36})[/tex]
Rewrite as perfect squares
[tex]y+\frac{8}{3}=\frac{3}{2}(x-\frac{4}{3})^{2}[/tex]
[tex]y=\frac{3}{2}(x-\frac{4}{3})^{2}-\frac{8}{3}[/tex]
In this problem we have
the vertex is the point [tex](\frac{4}{3},-\frac{8}{3})[/tex]
[tex]a=\frac{3}{2}[/tex]
therefore
the parabola open upward (vertex is a minimum)
The graph with its corresponding equation in the attached figure
case C) [tex]y=-x^{2}+2x+4[/tex]
convert to vertex form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]y-4=-(x^{2}-2x)[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side.
[tex]y-4-1=-(x^{2}-2x+1)[/tex]
[tex]y-5=-(x^{2}-2x+1)[/tex]
Rewrite as perfect squares
[tex]y-5=-(x-1)^{2}[/tex]
[tex]y=-(x-1)^{2}+5[/tex]
In this problem we have
the vertex is the point [tex](1,5)[/tex]
[tex]a=-1[/tex]
the y-intercept is [tex]4[/tex]
remember that the y-intercept is the value of y when the value of x is equal to zero
therefore
the parabola open downward (vertex is a maximum)
The graph with its corresponding equation is not in the attached figure
case D) [tex]y=\frac{1}{2} x^{2}+2x+3[/tex]
convert to vertex form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]y-3=\frac{1}{2} x^{2}+2x[/tex]
Factor the leading coefficient
[tex]y-3=\frac{1}{2}(x^{2}+4x)[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side.
[tex]y-3+2=\frac{1}{2}(x^{2}+4x+4)[/tex]
[tex]y-1=\frac{1}{2}(x^{2}+4x+4)[/tex]
Rewrite as perfect squares
[tex]y-1=\frac{1}{2}(x+2)^{2}[/tex]
[tex]y=\frac{1}{2}(x+2)^{2}+1[/tex]
In this problem we have
the vertex is the point [tex](-2,1)[/tex]
[tex]a=\frac{1}{2}[/tex]
therefore
the parabola open upward (vertex is a minimum)
The graph with its corresponding equation in the attached figure
case E) [tex]y=-2x^{2}+4x+3[/tex]
convert to vertex form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]y-3=-2x^{2}+4x[/tex]
Factor the leading coefficient
[tex]y-3=-2(x^{2}-2x)[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side.
[tex]y-3-2=-2(x^{2}-2x+1)[/tex]
[tex]y-5=-2(x^{2}-2x+1)[/tex]
Rewrite as perfect squares
[tex]y-5=-2(x-1)^{2}[/tex]
[tex]y=-2(x-1)^{2}+5[/tex]
In this problem we have
the vertex is the point [tex](1,5)[/tex]
[tex]a=-2[/tex]
the y-intercept is [tex]3[/tex]
remember that the y-intercept is the value of y when the value of x is equal to zero
therefore
the parabola open downward (vertex is a maximum)
The graph with its corresponding equation in the attached figure
case F) [tex]y=\frac{1}{2} x^{2}+3[/tex]
In this problem we have
the vertex is the point [tex](0,3)[/tex]
[tex]a=\frac{1}{2}[/tex]
therefore
the parabola open upward (vertex is a minimum)
The graph with its corresponding equation is not in the attached figure
therefore
the answer in the attached figure
