Quadratic Function and Parabola: Very Difficult Problems with Solutions

Solution:
Substituting the coordinates of the three points:
[tex]\left\{\begin{array}{l} 4a-2b+c=7 \\ a-b+c=3 \\ 9a+3b+c=7 \end{array}\right.[/tex]
Subtracting the first equation from the third gives [tex]5a+5b=0[/tex], i.e. [tex]b=-a[/tex]; then [tex]6a+c=7[/tex] and [tex]2a+c=3[/tex], so [tex]a=1[/tex], [tex]b=-1[/tex], [tex]c=1[/tex] and
[tex]f(x)=x^2-x+1[/tex]
The line [tex]BD[/tex] has slope [tex]\frac{7-3}{3+1}=1[/tex], so its equation is [tex]y=x+4[/tex], and [tex]BD=\sqrt{4^2+4^2}=4\sqrt{2}[/tex]. The diagonal [tex]BD[/tex] divides [tex]ABCD[/tex] into the triangles [tex]ABD[/tex] and [tex]CBD[/tex], and [tex]A[/tex] and [tex]C[/tex] lie on opposite sides of [tex]BD[/tex].
Triangle [tex]ABD[/tex] is fixed: [tex]AD[/tex] lies on the line [tex]y=7[/tex], [tex]AD=5[/tex] and the distance from [tex]B[/tex] to this line is [tex]4[/tex], so
[tex]S_{ABD}=\frac{1}{2}\times 5\times 4=10[/tex]
The point [tex]A[/tex] is above the line [tex]BD[/tex] (at [tex]x=-2[/tex] the line has [tex]y=2\lt 7[/tex]), so [tex]C[/tex] must be below it:
[tex]x^2-x+1\lt x+4\Leftrightarrow (x+1)(x-3)\lt 0\Leftrightarrow -1\lt x\lt 3[/tex]
Let [tex]C\left(t, t^2-t+1\right)[/tex] with [tex]-1\lt t\lt 3[/tex]. Its vertical distance to the line is [tex](t+4)-(t^2-t+1)=4-(t-1)^2[/tex], and since the line makes an angle of [tex]45^\circ[/tex] with the axes, the distance from [tex]C[/tex] to [tex]BD[/tex] is [tex]\frac{4-(t-1)^2}{\sqrt{2}}[/tex]. Hence
[tex]S_{CBD}=\frac{1}{2}\times 4\sqrt{2}\times \frac{4-(t-1)^2}{\sqrt{2}}=2\left(4-(t-1)^2\right)\le 8[/tex]
with equality only for [tex]t=1[/tex], i.e. for [tex]C(1, 1)[/tex] (the point where the tangent [tex]y=x[/tex] is parallel to [tex]BD[/tex]). For [tex]C(1, 1)[/tex] the segments [tex]AC[/tex] and [tex]BD[/tex] meet at [tex]\left(-\frac{1}{3}, \frac{11}{3}\right)[/tex], an interior point of both, so [tex]ABCD[/tex] is indeed a convex quadrilateral with diagonals [tex]AC[/tex] and [tex]BD[/tex]. The greatest area is
[tex]S_{ABCD}=S_{ABD}+S_{CBD}=10+8=18[/tex]

Solution:
Let [tex]A(x_1, 0)[/tex] and [tex]B(x_2, 0)[/tex], where [tex]x_1\ne x_2[/tex] are the roots of [tex]x^2+px+q=0[/tex], and [tex]C(0, q)[/tex]. Since [tex]q\ne 0[/tex], the point [tex]C[/tex] does not lie on the [tex]x[/tex]-axis, so [tex]A[/tex], [tex]B[/tex], [tex]C[/tex] are not collinear and exactly one circle passes through them. Let its equation be
[tex]x^2+y^2+Dx+Ey+F=0[/tex]
Its common points with the [tex]x[/tex]-axis ([tex]y=0[/tex]) satisfy [tex]x^2+Dx+F=0[/tex]. Since [tex]A[/tex] and [tex]B[/tex] lie on the circle, this equation has the distinct roots [tex]x_1[/tex] and [tex]x_2[/tex], hence
[tex]x^2+Dx+F=(x-x_1)(x-x_2)=x^2+px+q\Rightarrow D=p,\quad F=q[/tex]
The point [tex]C(0, q)[/tex] lies on the circle, and [tex]q\ne 0[/tex]:
[tex]q^2+Eq+q=0\Rightarrow q+E+1=0\Rightarrow E=-(q+1)[/tex]
So the circle through [tex]A[/tex], [tex]B[/tex], [tex]C[/tex] is
[tex]x^2+y^2+px-(q+1)y+q=0[/tex]
Substituting the coordinates of [tex]P(0, 1)[/tex]:
[tex]0+1+0-(q+1)+q=0[/tex]
which is true. Hence [tex]P[/tex] also lies on this circle, i.e. the points [tex]A[/tex], [tex]B[/tex], [tex]C[/tex] and [tex]P[/tex] lie on one circle. (If [tex]q=1[/tex], then [tex]C=P[/tex] and the statement is obvious.)

Solution:
Let [tex]g(x)=ax^2+bx+c[/tex]. For every such function
[tex]g(4)-3g(3)+3g(2)-g(1)=a(16-27+12-1)+b(4-9+6-1)+c(1-3+3-1)=0[/tex]
hence
[tex]g(4)=g(1)-3g(2)+3g(3)[/tex]
The numbers [tex]1[/tex], [tex]2[/tex], [tex]3[/tex] belong to [tex]\left[1, 3\right][/tex], so [tex]|g(1)|\le 1[/tex], [tex]|g(2)|\le 1[/tex], [tex]|g(3)|\le 1[/tex]. Therefore
[tex]g(4)\le |g(1)|+3|g(2)|+3|g(3)|\le 1+3+3=7[/tex]
The bound is attained: [tex]g(x)=2(x-2)^2-1=2x^2-8x+7[/tex] satisfies [tex]-1\le g(x)\le 1[/tex] on [tex]\left[1, 3\right][/tex] (its least value there is [tex]g(2)=-1[/tex] and its greatest is [tex]g(1)=g(3)=1[/tex]), and [tex]g(4)=7[/tex].

Solution:
First, [tex]a\ne 0[/tex]: for [tex]a=0[/tex] the condition would become [tex](b-c)^2\lt 0[/tex], which is impossible. Take
[tex]x_0=\frac{c-b}{2a},\qquad ax_0=\frac{c-b}{2}[/tex]
Then
[tex]f(x_0)=x_0^2+\frac{c-b}{2}+b=x_0^2+\frac{b+c}{2},\qquad g(x_0)=x_0^2-\frac{c-b}{2}+c=x_0^2+\frac{b+c}{2}[/tex]
Hence, by the condition,
[tex]f(x_0)=g(x_0)=\frac{(b-c)^2}{4a^2}+\frac{b+c}{2}=\frac{(b-c)^2+2a^2(b+c)}{4a^2}\lt 0[/tex]
Each of the functions [tex]f[/tex] and [tex]g[/tex] has leading coefficient [tex]1\gt 0[/tex] (its parabola opens upward) and takes a negative value at [tex]x_0[/tex]. Hence its least value [tex]-\frac{D}{4}[/tex] is negative, i.e. its discriminant [tex]D[/tex] is positive, and its parabola crosses the [tex]x[/tex]-axis at two points, one on each side of [tex]x_0[/tex]. Therefore each of the equations [tex]f(x)=0[/tex] and [tex]g(x)=0[/tex] has two distinct real roots.

Solution:
Let [tex]f(x)=x^2-2mx+8[/tex]. Then
[tex]\frac{D}{4}=m^2-8[/tex]
If [tex]m^2\le 8[/tex], i.e. [tex]|m|\le 2\sqrt{2}[/tex], then [tex]f(x)\ge 0[/tex] for every real [tex]x[/tex], in particular for every integer [tex]x[/tex].
Let [tex]m^2\gt 8[/tex]. Then [tex]f[/tex] has roots [tex]x_1\lt x_2[/tex] and [tex]f(x)\lt 0[/tex] exactly for [tex]x\in\left(x_1, x_2\right)[/tex], so the condition means that this open interval contains no integer. The expression does not change if [tex]m[/tex] and [tex]x[/tex] are replaced by [tex]-m[/tex] and [tex]-x[/tex], and [tex]-x[/tex] is an integer exactly when [tex]x[/tex] is; so the set of suitable [tex]m[/tex] is symmetric about [tex]0[/tex] and it suffices to consider [tex]m\gt 2\sqrt{2}[/tex]. By Vieta's formulas
[tex]x_1+x_2=2m\gt 0,\quad x_1x_2=8\Rightarrow 0\lt x_1\lt 2\sqrt{2}\lt x_2[/tex]
Since [tex]2\lt 2\sqrt{2}\lt 3[/tex], the interval [tex]\left(x_1, x_2\right)[/tex] contains no integer if and only if [tex]x_1\ge 2[/tex] and [tex]x_2\le 3[/tex]. As [tex]x_2\gt 2[/tex], the condition [tex]x_1\ge 2[/tex] is equivalent to [tex]f(2)\ge 0[/tex]; as [tex]x_1\lt 3[/tex], the condition [tex]x_2\le 3[/tex] is equivalent to [tex]f(3)\ge 0[/tex]:
[tex]f(2)=12-4m\ge 0\Rightarrow m\le 3,\qquad f(3)=17-6m\ge 0\Rightarrow m\le\frac{17}{6}[/tex]
Since [tex]\left(\frac{17}{6}\right)^2=\frac{289}{36}\gt 8[/tex], we get [tex]2\sqrt{2}\lt m\le\frac{17}{6}[/tex]. For [tex]m=\frac{17}{6}[/tex] the roots are [tex]\frac{8}{3}[/tex] and [tex]3[/tex], and [tex]f(3)=0[/tex] is allowed, so this endpoint is included. By symmetry also [tex]-\frac{17}{6}\le m\lt -2\sqrt{2}[/tex]. Together with [tex]|m|\le 2\sqrt{2}[/tex] we obtain
[tex]m\in\left[-\frac{17}{6}, \frac{17}{6}\right][/tex]

Solution:
Complete the square:
[tex]f(x)=(2x-a)^2-2a+2[/tex]
The parabola opens upward, its vertex is at [tex]x_0=\frac{a}{2}[/tex] and [tex]f(x_0)=2-2a[/tex]. We consider the position of [tex]x_0[/tex] with respect to [tex]\left[0, 2\right][/tex].
Case [tex]1[/tex]: [tex]\frac{a}{2}\lt 0[/tex], i.e. [tex]a\lt 0[/tex]. Then [tex]f[/tex] is increasing on [tex]\left[0, 2\right][/tex] and its least value is [tex]f(0)[/tex]:
[tex]a^2-2a+2=3\Rightarrow a^2-2a-1=0\Rightarrow a=1\pm\sqrt{2}[/tex]
Only [tex]a=1-\sqrt{2}\lt 0[/tex] satisfies the case condition.
Case [tex]2[/tex]: [tex]0\le\frac{a}{2}\le 2[/tex], i.e. [tex]0\le a\le 4[/tex]. The least value is attained at the vertex:
[tex]2-2a=3\Rightarrow a=-\frac{1}{2}\notin\left[0, 4\right][/tex]
so this case gives no solution.
Case [tex]3[/tex]: [tex]\frac{a}{2}\gt 2[/tex], i.e. [tex]a\gt 4[/tex]. Then [tex]f[/tex] is decreasing on [tex]\left[0, 2\right][/tex] and its least value is [tex]f(2)[/tex]:
[tex]16-8a+a^2-2a+2=3\Rightarrow a^2-10a+15=0\Rightarrow a=5\pm\sqrt{10}[/tex]
Since [tex]5-\sqrt{10}\lt 2\lt 4[/tex], only [tex]a=5+\sqrt{10}[/tex] satisfies the case condition. Hence
[tex]a\in\left\{1-\sqrt{2}, 5+\sqrt{10}\right\}[/tex]

Solution:
The axis of symmetry of the parabola is the line [tex]x=1[/tex], so
[tex]f(x)=a(x-1)^2+k,\quad a\lt 0[/tex]
The parabola is tangent to the (non-vertical) line exactly when the equation [tex]a(x-1)^2+k=-2x-3[/tex], i.e. [tex]ax^2+(2-2a)x+a+k+3=0[/tex], has a double root:
[tex]\frac{D}{4}=(1-a)^2-a(a+k+3)=1-5a-ak=0\Rightarrow k=\frac{1-5a}{a}=\frac{1}{a}-5[/tex]
The parabola meets the [tex]y[/tex]-axis at the point with ordinate
[tex]f(0)=a+k=a+\frac{1}{a}-5[/tex]
For [tex]a\lt 0[/tex], by the AM-GM inequality,
[tex]a+\frac{1}{a}=-\left(|a|+\frac{1}{|a|}\right)\le -2[/tex]
with equality only for [tex]a=-1[/tex]. So the greatest value of [tex]f(0)[/tex] is [tex]-2-5=-7[/tex], attained for [tex]a=-1[/tex], [tex]k=-6[/tex]:
[tex]f(x)=-(x-1)^2-6=-x^2+2x-7[/tex]
Check: [tex]-x^2+2x-7=-2x-3\Leftrightarrow (x-2)^2=0[/tex], so the parabola touches the line at [tex](2, -7)[/tex].