Quadratic Function and Parabola: Problems with Solutions

Solution:
The coordinates of the vertex are
[tex]x_V=-\frac{b}{2a}=-\frac{4}{2}=-2,\quad y_V=(-2)^2+4\times (-2)+5=1[/tex]
Indeed, [tex]y=(x+2)^2+1[/tex], so the vertex is [tex]V(-2, 1)[/tex]. The distance from a point to the [tex]y[/tex]-axis (the line [tex]x=0[/tex]) equals the absolute value of its [tex]x[/tex]-coordinate:
[tex]d=|x_V|=|-2|=2[/tex]
(The number [tex]1[/tex] is the distance from the vertex to the [tex]x[/tex]-axis, not to the [tex]y[/tex]-axis.)

Solution:
The least value [tex]0[/tex] is attained at [tex]x=1[/tex], so [tex](1, 0)[/tex] is the vertex of the parabola and the parabola opens upward:
[tex]f(x)=a(x-1)^2,\quad a\gt 0[/tex]
From [tex]f(-1)=8[/tex] we get
[tex]a\times (-1-1)^2=4a=8\Rightarrow a=2\gt 0[/tex]
Hence [tex]f(x)=2(x-1)^2=2x^2-4x+2[/tex] and
[tex]f(-2)=2\times (-3)^2=18[/tex]

Solution:
Since [tex]a=b+6[/tex], the product is a quadratic function of [tex]b[/tex] with positive leading coefficient:
[tex]ab=b(b+6)=b^2+6b=(b+3)^2-9\ge -9[/tex]
Equality holds at the vertex [tex]b=-3[/tex], i.e. for [tex]a=3[/tex], [tex]b=-3[/tex]. Hence the least value of [tex]ab[/tex] is [tex]-9[/tex].

Solution:
Complete the square:
[tex]f(x)=-x^2-2x+19=-(x+1)^2+20[/tex]
The parabola opens downward and its vertex is [tex](-1, 20)[/tex]. Since [tex]-1\in\left[-2, 1\right][/tex], the greatest value is [tex]f(-1)=20[/tex]. The function increases on [tex]\left[-2, -1\right][/tex] and decreases on [tex]\left[-1, 1\right][/tex], so the least value is attained at an endpoint:
[tex]f(-2)=19+4-4=19,\quad f(1)=19-2-1=16[/tex]
So the least value is [tex]16[/tex], and the required sum is
[tex]20+16=36[/tex]

Solution:
Let one side of the rectangle be [tex]x[/tex] ft. The half-perimeter is [tex]4[/tex] ft, so the other side is [tex]4-x[/tex] ft and [tex]0\lt x\lt 4[/tex]. The area is
[tex]S(x)=x(4-x)=-x^2+4x=4-(x-2)^2\le 4[/tex]
The parabola opens downward and its vertex [tex]x=2[/tex] lies in [tex]\left(0, 4\right)[/tex], so the greatest area is [tex]S(2)=4[/tex]: the window is a square with side [tex]2[/tex] ft, and its area is [tex]4[/tex] square feet.

Solution:
Write the function in vertex form:
[tex]y=5+2x-x^2=-(x-1)^2+6[/tex]
A translation does not change the leading coefficient (the shape and the opening of the parabola), only the position of the vertex. With the new vertex [tex](4, 6)[/tex] we get
[tex]y=-(x-4)^2+6=-x^2+8x-16+6=-x^2+8x-10[/tex]
Equivalently, a shift by [tex]3[/tex] units to the right replaces [tex]x[/tex] by [tex]x-3[/tex]:
[tex]5+2(x-3)-(x-3)^2=-x^2+8x-10[/tex]
(Replacing [tex]x[/tex] by [tex]x+3[/tex] would move the vertex to [tex](-2, 6)[/tex], adding [tex]3[/tex] to the function would move it to [tex](1, 9)[/tex], and [tex]y=x^2-8x+22[/tex] opens upward.)

Solution:
The graph has no common points with the [tex]x[/tex]-axis, so the discriminant is negative and the parabola lies entirely above the axis (when [tex]a\gt 0[/tex]) or entirely below it (when [tex]a\lt 0[/tex]).
The point [tex](1, f(1))[/tex] of the graph lies above the axis because [tex]f(1)\gt 0[/tex], hence the whole parabola lies above the axis. Therefore it opens upward and [tex]f(x)\gt 0[/tex] for every [tex]x[/tex]; in particular, for [tex]x=0[/tex]:
[tex]a\gt 0,\quad c=f(0)\gt 0[/tex]