Quadratic Function and Parabola: Difficult Problems with Solutions

Solution:
The constant term of [tex]f(x+k)[/tex] is its value at [tex]x=0[/tex], i.e. [tex]f(k)[/tex]. So [tex]k[/tex] must satisfy
[tex]4k^2+15k-4=0,\quad D=225+64=289,\quad k=\frac{-15\pm 17}{8}[/tex]
i.e. [tex]k=\frac{1}{4}[/tex] or [tex]k=-4[/tex]. The smaller value is [tex]k_1=-4[/tex], and
[tex]f(x-4)=4(x-4)^2+15(x-4)-4=4x^2-32x+64+15x-60-4=4x^2-17x=x(4x-17)[/tex]
Its zeros are [tex]x=0[/tex] and [tex]x=\frac{17}{4}[/tex], so the intersection points are [tex](0, 0)[/tex] and [tex]\left(\frac{17}{4}, 0\right)[/tex].
Check: the graph of [tex]f(x-4)[/tex] is the graph of [tex]f[/tex] shifted by [tex]4[/tex] units to the right, so the zeros [tex]-4[/tex] and [tex]\frac{1}{4}[/tex] of [tex]f[/tex] become [tex]0[/tex] and [tex]\frac{17}{4}[/tex].

Solution:
Since [tex]x=-1[/tex] is the axis of symmetry, [tex]f(x)=a(x+1)^2+n[/tex] with [tex]a\ne 0[/tex]. From [tex]f(0)=a+n=1[/tex] we get [tex]n=1-a[/tex], so
[tex]f(x)=a(x+1)^2+1-a[/tex]
The interval [tex]\left[2, 4\right][/tex] lies to the right of the axis [tex]x=-1[/tex], so [tex]f[/tex] is monotonic on it. If [tex]a\gt 0[/tex], [tex]f[/tex] increases on [tex]\left[2, 4\right][/tex] and its least value is
[tex]f(2)=9a+1-a=8a+1=-7\Rightarrow a=-1[/tex]
which contradicts [tex]a\gt 0[/tex]. If [tex]a\lt 0[/tex], [tex]f[/tex] decreases on [tex]\left[2, 4\right][/tex] and its least value is
[tex]f(4)=25a+1-a=24a+1=-7\Rightarrow a=-\frac{1}{3}\lt 0[/tex]
Hence
[tex]f(x)=-\frac{1}{3}(x+1)^2+\frac{4}{3}=-\frac{1}{3}x^2-\frac{2}{3}x+1[/tex]
Check: [tex]f(0)=1[/tex], [tex]f(4)=-7[/tex] and [tex]f(2)=-\frac{5}{3}\gt -7[/tex].

Solution:
The leading coefficient is [tex]-3\lt 0[/tex], so the parabola opens downward: [tex]f[/tex] increases on [tex]\left(-\infty, x_0\right][/tex] and decreases on [tex]\left[x_0, +\infty\right)[/tex], where
[tex]x_0=-\frac{-a}{2\times (-3)}=-\frac{a}{6}[/tex]
If [tex]x_0\le -1[/tex], then [tex]\left(-1, 2\right)\subset\left[x_0, +\infty\right)[/tex] and [tex]f[/tex] is decreasing on the interval. If [tex]x_0\gt -1[/tex], then [tex]f[/tex] increases on the non-empty part [tex]\left(-1, \min\left\{x_0, 2\right\}\right)[/tex] of the interval, so it is not decreasing there. Hence the condition is
[tex]-\frac{a}{6}\le -1\Leftrightarrow a\ge 6[/tex]
For [tex]a=6[/tex] the vertex is exactly at [tex]x=-1[/tex] and [tex]f[/tex] decreases on [tex]\left[-1, +\infty\right)[/tex], so [tex]a=6[/tex] is included. Hence [tex]a\in\left[6, +\infty\right)[/tex].

Solution:
The abscissas of the common points are the solutions of [tex]x^2-2x-3=ax^2[/tex] (each solution gives exactly one common point), i.e. of the equation
[tex](1-a)x^2-2x-3=0[/tex]
If [tex]a=1[/tex], the equation is linear, [tex]-2x-3=0[/tex], with the single solution [tex]x=-\frac{3}{2}[/tex]; there is exactly one common point [tex]\left(-\frac{3}{2}, \frac{9}{4}\right)[/tex].
If [tex]a\ne 1[/tex], the equation is quadratic and has exactly one root if and only if its discriminant is zero:
[tex]\frac{D}{4}=1+3(1-a)=4-3a=0\Leftrightarrow a=\frac{4}{3}[/tex]
Then the equation is [tex]-\frac{1}{3}x^2-2x-3=0\Leftrightarrow (x+3)^2=0[/tex], and the parabolas touch at [tex](-3, 12)[/tex]. For all other values of [tex]a[/tex] there are two common points or none.
Hence [tex]a\in\left\{1, \frac{4}{3}\right\}[/tex]. The case [tex]a=1[/tex] is easy to lose if one only requires [tex]D=0[/tex].

Solution:
The abscissas of the common points satisfy [tex]f(x)=g(x)[/tex]:
[tex]7x^2-4x-3=0\Leftrightarrow (x-1)(7x+3)=0\Leftrightarrow x=1\text{ or }x=-\frac{3}{7}[/tex]
Since [tex]f(1)=-1[/tex] and [tex]f\left(-\frac{3}{7}\right)=\frac{18}{49}+\frac{6}{7}-1=\frac{11}{49}[/tex], the points are [tex](1, -1)[/tex] and [tex]\left(-\frac{3}{7}, \frac{11}{49}\right)[/tex]. The slope of the line through them is
[tex]a=\frac{\frac{11}{49}+1}{-\frac{3}{7}-1}=\frac{\frac{60}{49}}{-\frac{10}{7}}=-\frac{6}{7}[/tex]
and from [tex]-1=a\times 1+b[/tex] we get [tex]b=-1+\frac{6}{7}=-\frac{1}{7}[/tex].
A shortcut: at a common point [tex]y=f(x)=g(x)[/tex], so
[tex]7y=5f(x)+2g(x)=10x^2-10x-5-10x^2+4x+4=-6x-1[/tex]
The [tex]x^2[/tex] terms cancel, hence both points lie on the line [tex]y=-\frac{6}{7}x-\frac{1}{7}[/tex], i.e. [tex]a=-\frac{6}{7}[/tex], [tex]b=-\frac{1}{7}[/tex].

Solution:
With [tex]x[/tex] new trees ([tex]x[/tex] a non-negative integer) there are [tex]30+x[/tex] trees, each yielding [tex]25-0.6x[/tex] pounds. The harvest is
[tex]R(x)=(30+x)(25-0.6x)=-0.6x^2+7x+750[/tex]
This parabola opens downward and its vertex is at
[tex]x_0=\frac{7}{2\times 0.6}=\frac{35}{6}\approx 5.83[/tex]
which is not an integer. [tex]R[/tex] increases for [tex]x\le\frac{35}{6}[/tex], decreases for [tex]x\ge\frac{35}{6}[/tex] and is symmetric about [tex]x_0[/tex], so among the integers the maximum is at the integer closest to [tex]\frac{35}{6}[/tex], namely [tex]x=6[/tex] (its distance to [tex]x_0[/tex] is [tex]\frac{1}{6}[/tex], while for [tex]x=5[/tex] it is [tex]\frac{5}{6}[/tex]). Check:
[tex]R(5)=35\times 22=770, \quad R(6)=36\times 21.4=770.4, \quad R(7)=37\times 20.8=769.6[/tex]
Hence [tex]6[/tex] new trees should be planted.

Solution:
Since [tex]f(x)=(x-1)^2-1[/tex], the equation [tex]f(x)=t[/tex] has two solutions [tex]x=1\pm\sqrt{t+1}[/tex] if [tex]t\gt -1[/tex], one solution [tex]x=1[/tex] if [tex]t=-1[/tex] (the vertex) and none if [tex]t\lt -1[/tex]; different values of [tex]t[/tex] give different solutions. We work backwards.
Let [tex]y_3=f(f(f(c)))[/tex]. Then [tex]f(y_3)=3\Leftrightarrow (y_3-1)^2=4[/tex], so [tex]y_3\in\left\{3, -1\right\}[/tex].
Let [tex]y_2=f(f(c))[/tex]. From [tex]f(y_2)=3[/tex] we get [tex]y_2\in\left\{3, -1\right\}[/tex], and from [tex]f(y_2)=-1[/tex] we get [tex]y_2=1[/tex], so [tex]y_2\in\left\{3, -1, 1\right\}[/tex].
Let [tex]y_1=f(c)[/tex]. From [tex]f(y_1)=3[/tex] we get [tex]y_1\in\left\{3, -1\right\}[/tex], from [tex]f(y_1)=-1[/tex] we get [tex]y_1=1[/tex], and [tex]f(y_1)=1[/tex] gives [tex](y_1-1)^2=2[/tex], i.e. [tex]y_1=1\pm\sqrt{2}[/tex]. So [tex]y_1[/tex] takes five distinct values:
[tex]y_1\in\left\{3, -1, 1, 1+\sqrt{2}, 1-\sqrt{2}\right\}[/tex]
Finally, we solve [tex]f(c)=y_1[/tex]: for [tex]y_1=-1[/tex] there is one solution [tex]c=1[/tex], and each of the other four values is greater than [tex]-1[/tex] (note that [tex]1-\sqrt{2}\approx -0.41\gt -1[/tex]), so each gives two solutions. Hence the number of distinct numbers [tex]c[/tex] is
[tex]1+4\times 2=9[/tex]

Solution:
We have [tex]f(-1)=a-b+c[/tex], [tex]f(1)=a+b+c[/tex] and [tex]f(3)=9a+3b+c[/tex]. Then
[tex]f(-1)-2f(1)+f(3)=(a-2a+9a)+(-b-2b+3b)+(c-2c+c)=8a[/tex]
From the conditions, [tex]f(-1)\gt -4[/tex], [tex]-2f(1)\gt 0[/tex] and [tex]f(3)\gt 5[/tex]. Adding these inequalities gives
[tex]8a=f(-1)-2f(1)+f(3)\gt -4+0+5=1\Rightarrow a\gt\frac{1}{8}\gt 0[/tex]
Thus [tex]a\gt 0[/tex], which completes the proof.