$bold{Concept:}$ A quadratic function of the form

$f(x)=ax^2+bx+c$ $;;;;;$ or $;;;;;$ $f(x)=a(x-h)^2+k$

will have a maximum if it opens downwards $(a0)$

$bold{Solution:}$ Rewrite each expressions in the form $f(x)=ax^2+bx+c$ $;;$ or $;;$ $f(x)=a(x-h)^2+k$ and determine the sign of $a$.

a) $y=-x^2+7ximplies a=-1$

b) $f(x)=3(x-1)^2-4implies a=3$

c) $f(x)=-4(x+2)(x-3)=-4(x^2-x-6)=-4x^2+4x+24$
$implies a=-4$

d) $g(x)=4x^2+3x-5implies a=4$

Therefore, the answers are options a) and c)

$bold{Concept:}$ The $bold{vertex}$ $(h,k)$ of a parabola is the turning point which corresponds to either maximum or minimum value.

If the parabola opens upward, the minimum value is $k$.

If the parabola opens downward, the maximum value is $k$.

a) The parabola opens upward with vertex $(-5,-2)$, so the minimum value is $-2$

b) The parabola opens downward with vertex $(4,8)$, so the maximum value is $8$

a) vertex$(-5,-2)$, minimum value: $-2$

b) vertex$(4,8)$, maximum value: $8$

a) In the vertex form, which is given by $f(x)=a(x-h)^2+k,$ the given equation, $y=-4(x+1)^2+6
,$ is equivalent to

$$
begin{align*}
y=-4[x-(-1)]^2+6
.end{align*}
$$

Since the maximum/minimum point of the quadratic function in the form $f(x)=a(x-h)^2+k,$ is $y=k,$ then the maximum/minimum point of the equation above is

$$
begin{align*}
y=6
.end{align*}
$$

b) In the vertex form, which is given by $f(x)=a(x-h)^2+k,$ the given equation, $f(x)=(x-5)^2
,$ is equivalent to

$$
begin{align*}
f(x)=(x-5)^2+0
.end{align*}
$$

Since the maximum/minimum point of the quadratic function in the form $f(x)=a(x-h)^2+k,$ is $y=k,$ then the maximum/minimum point of the equation above is

$$
begin{align*}
y=0
.end{align*}
$$

c) In the vertex form, which is given by $f(x)=a(x-h)^2+k,$ the given equation, $f(x)=-2x(x-4)
,$ is equivalent to

$$
begin{align*}
f(x)&=-2x^2+8x
\
f(x)&=-2(x^2-4x)
\
f(x)&=-2(x^2-4x+4)+2(4)
\
f(x)&=-2(x-2)^2+8
.end{align*}
$$

Since the maximum/minimum point of the quadratic function in the form $f(x)=a(x-h)^2+k,$ is $y=k,$ then the maximum/minimum point of the equation above is

$$
begin{align*}
y=8
.end{align*}
$$

d) In the vertex form, which is given by $f(x)=a(x-h)^2+k,$ the given equation, $g(x)=2x^2-7
,$ is equivalent to

$$
begin{align*}
g(x)&=2(x-0)^2-7
.end{align*}
$$

Since the maximum/minimum point of the quadratic function in the form $f(x)=a(x-h)^2+k,$ is $y=k,$ then the maximum/minimum point of the equation above is

$$
begin{align*}
y=-7
.end{align*}
$$

a) $y=6$

b) $y=0$

c) $y=8$

d) $y=-7$

a) In the vertex form, which is given by $f(x)=a(x-h)^2+k,$ the given equation, $y=x^2-4x-1
,$ is equivalent to

$$
begin{align*}
y&=(x^2-4x)-1
\
y&=(x^2-4x+4)-1+4
\
y&=(x-2)^2+3
.end{align*}
$$

Since the maximum/minimum point of the quadratic function in the form $f(x)=a(x-h)^2+k,$ is $y=k,$ then the maximum/minimum point of the equation above is

$$
begin{align*}
y=3
.end{align*}
$$

b) In the vertex form, which is given by $f(x)=a(x-h)^2+k,$ the given equation, $f(x)=x^2-8x+12
,$ is equivalent to

$$
begin{align*}
f(x)&=(x^2-8x)+12
\
f(x)&=(x^2-8x+16)+12-16
\
f(x)&=(x-4)^2-4
.end{align*}
$$

Since the maximum/minimum point of the quadratic function in the form $f(x)=a(x-h)^2+k,$ is $y=k,$ then the maximum/minimum point of the equation above is

$$
begin{align*}
y=-4
.end{align*}
$$

c) In the vertex form, which is given by $f(x)=a(x-h)^2+k,$ the given equation, $y=2x^2+12x
,$ is equivalent to

$$
begin{align*}
y&=2(x^2+6x)
\
y&=2(x^2+6x+9)-2(9)
\
y&=2(x+3)^2-18
.end{align*}
$$

Since the maximum/minimum point of the quadratic function in the form $f(x)=a(x-h)^2+k,$ is $y=k,$ then the maximum/minimum point of the equation above is

$$
begin{align*}
y=-18
.end{align*}
$$

d) In the vertex form, which is given by $f(x)=a(x-h)^2+k,$ the given equation, $y=-3x^2-12x+15
,$ is equivalent to

$$
begin{align*}
y&=(-3x^2-12x)+15
\
y&=-3(x^2+4x)+15
\
y&=-3(x^2+4x+4)+15+3(4)
\
y&=-3(x+2)^2+15+12
\
y&=-3(x+2)^2+27
.end{align*}
$$

Since the maximum/minimum point of the quadratic function in the form $f(x)=a(x-h)^2+k,$ is $y=k,$ then the maximum/minimum point of the equation above is

$$
begin{align*}
y=27
.end{align*}
$$

e) In the vertex form, which is given by $f(x)=a(x-h)^2+k,$ the given equation, $y=3x(x-2)+5
,$ is equivalent to

$$
begin{align*}
y&=3x^2-6x+5
\
y&=(3x^2-6x)+5
\
y&=3(x^2-2x)+5
\
y&=3(x^2-2x+1)+5-3(1)
\
y&=3(x-1)^2+5-3
\
y&=3(x-1)^2+2
.end{align*}
$$

Since the maximum/minimum point of the quadratic function in the form $f(x)=a(x-h)^2+k,$ is $y=k,$ then the maximum/minimum point of the equation above is

$$
begin{align*}
y=2
.end{align*}
$$

e) In the vertex form, which is given by $f(x)=a(x-h)^2+k,$ the given equation, $g(x)=-2(x+1)^2-5
,$ is equivalent to

$$
begin{align*}
g(x)&=-2[x-(-1)]^2-5
.end{align*}
$$

Since the maximum/minimum point of the quadratic function in the form $f(x)=a(x-h)^2+k,$ is $y=k,$ then the maximum/minimum point of the equation above is

$$
begin{align*}
y=-5
.end{align*}
$$

a) $y=3$

b) $y=-4$

c) $y=-18$

d) $y=27$

e) $y=2$

f) $y=-5$

$bold{Concept:}$ If $p(x)$ is the demand function and $x$ is the number of items sold, then the revenue function is

Revenue function: $R(x)= [p(x)](x)$

To obtain the maximum revenue, solve for zeroes of the quadratic function. The maximum value corresponds to the revenue when $x$ is halfway between the two zeroes (at the axis of symmetry).

a) Revenue function= $[p(x)](x)=(-x+5)(x)=-x^2+5x$

zeroes: $(-x+5)=0implies x=5$ and $x=0$

The revenue is maximum when $x=dfrac{5+0}{2}=2.5$

$R(2.5) = -left(2.5right)^2+5left(2.5right)=6.25$

Since $R$ has units of thousand dollars, the maximum revenue is $$ 6,250$.

b) Revenue function: $R(x)=[p(x)](x)=(-4x+12)(x)=-4x^2+12x$

zeroes: $(-4x+12)=0implies x=dfrac{12}{4}=3$ and $x=0$

The revenue is maximum when $x=dfrac{3+0}{2}=1.5$

$R(1.5) = -4(1.5)^2+12(1.5)=9$

The maximum revenue is $$9000$

c) Revenue function : $R(x)=[p(x)](x)=(-0.6x+15)(x)=-0.6x^2+15x$

zeroes: $(-0.6x+15)=0implies x=dfrac{15}{0.6}=25$ and $x=0$

The revenue is maximum when $x=dfrac{25+0}{2}=12.5$

$R(12.5)=-0.6(12.5)^2+15(12.5)=93.75$

The maximum revenue is $$9375$

d) Revenue function : $R(x)=[p(x)](x)=(-1.2x+4.8)(x)=-1.2x^2+4.8x$

zeroes: $(-1.2x+4.8)=0implies x=dfrac{4.8}{1.2}=4$ and $x=0$

The revenue is maximum when $x=dfrac{4+0}{2}=2$

$R(2)=-1.2(2)^2+4.8(2)=4.8$

The maximum revenue is $$4,800$

a) i) $R(x)=-x^2+5x$ ; ii) maximum revenue: $$6,250$

b) i) $R(x)=-4x^2+12x$ ; ii) maximum revenue: $$9,000$

c) i) $R(x)=-0.6x^2+15x$ ; ii) maximum revenue: $$93,750$

d) i) $R(x)=-1.2x^2+4.8x$ ; ii) maximum revenue: $$4,800$

a) Using a graphing software, the graph of the given function, $f(x)=2x^2-6.5x+3.2
,$ is shown below.

Based on the graph below, the maximum/minimum value is $-2.08
.$

a) Using a graphing software, the graph of the given function, $f(x)=-3.6x^2+4.8x
,$ is shown below.

Based on the graph below, the maximum/minimum value is $1.6
.$

a) $-2.08$

b) $1.6$

$bold{Concept:}$ If the revenue function is $R(x)$, the cost function is $C(x)$, then the profit function $P(x)$ is

Profit = Revenue $-$ Cost

$P(x)=R(x)-C(x)$

The value of $x$ that maximizes the profit can be obtained by rewriting in the form

$R(x)=ax^2+bx+c$

This is the value at the axis of symmetry which is

$x=-dfrac{b}{2a}$

$bold{Solution}$

a) $P(x)=R(x)-C(x)$

$P(x)=-x^2+24x-(12x+28)$

$P(x)=-x^2+24x-12x-28$

profit function: $P(x)=-x^2+12x-28$

$a=-1$, $b=12$

$x=-dfrac{b}{2a}=-dfrac{12}{2(-1)}=6$

The maximum profit occurs when $x=6$

b) $P(x)=R(x)=C(x)$

$P(x)=-2x^2+32x-(14x+45)$

$P(x)=-2x^2+32x-14x-45$

$P(x)=-2x^2+18x-45$

$a=-2$ , $b=18$

$x=-dfrac{b}{2a}=-dfrac{18}{2(-2)}=4.5$

The maximum profit occurs when $x=4.5$

c) $P(x)=R(x)-C(x)$

$P(x)=-3x^2+26x-(8x+18)$

$P(x)=(-3x^2+26x)-(8x+18)$

$P(x)=-3x^2+26x-8x-18$

$P(x)=-3x^2+18x-18$

$a=-3$, $b=18$

$x=-dfrac{b}{2a}=-dfrac{18}{2(-3)}=3$

The maximum profit occurs when $x=3$

d) $P(x)=R(x)-C(x)$

$P(x)=-2x^2+25x-(3x+17)$

$P(x)=-2x^2+25x-3x-17$

$P(x)=-2x^2+22x-17$

$a=-2$, $b=22$

$x=-dfrac{b}{2a}=-dfrac{22}{2(-2)}=5.5$

The maximum profit occurs when $x=5.5$

a) i) $P(x)=-x^2+12x-28$ ; ii) $x=6$

b) i) $P(x)=-2x^2+18x-45$ ; ii) $x=4.5$

c) i) $P(x)=-3x^2+18x-18$ ; ii) $x=3$

d) i) $P(x)=-2x^2+22x-17$ ; ii) $x=5.5$

$bold{Concept:}$ For a general quadratic function of the form

$f(x)=ax^2+bx+c$

The value of $x$ that maximizes or minimizes $f(x)$ is

$x=-dfrac{b}{2a}$.

The maximum or minimum value of $f(x)$ is the value of $fleft(-dfrac{b}{2a}right)$.

$bold{Solution:}$

a) From the given equation, $h(t)=-5t^2+20t+50$

$a=-5$, $b=20$ , $c=20$

The value of $t$ when $h(t)$ is maximum is

$t=-dfrac{b}{2a}=-dfrac{20}{2(-5)}=2$

The maximum height is

$h(2)=-5(2)^2+20(2)+50=70$

The maximum height is $70$ m.

b) The time it takes to reach the maximum height is
obtained in part(a) as $t=2$

c) The height of the rooftop is the value of $h(t)$ when $t=0$

$h(0)=-5(0)^2+20(0)+50$

$h(0)=50$

The rooftop is 50 meters high.

a) $70$ m

b) $t=2$ s

c) 50 m

$bold{Concept:}$ For a general quadratic function of the form

$f(x)=ax^2+bx+c$

The value of $x$ which maximizes or minimizes $f(x)$ is

$x=-dfrac{b}{2a}$.

The maximum or minimum value of $f(x)$ is the value of $fleft(-dfrac{b}{2a}right)$.

$bold{Solution:}$

a) From the given equation, $C(x)=0.28x^2-0.7x+1$

$a=0.28$, $b=-0.7$ , $c=1$

The value of $x$ when $C(x)$ is minimum is

$x=-dfrac{b}{2a}=-dfrac{-0.7}{2(0.28)}=1.25$

The minimum value is

$C(1.25)=0.28(1.25)^2-0.7(1.25)+1=0.5625$

Since the unit of $C(x)$ is million dollars, the minimum production cost is

$0.5625$ million dollars or $$562,500$

$$
$562,500
$$

Let $y=3x^2-6x+5
.$ In the form $y=a(x-h)^2+k,$ then

$$
begin{align*}
y&=(3x^2-6x)+5
\
y&=3(x^2-2x)+5
\
y&=3(x^2-2x+1)+5-3(1)
\
y&=3(x^2-2x+1)+5-3
\
y&=3(x-1)^2+2
.end{align*}
$$

Since $a>0$ ($a=3$), then the graph is a parabola that opens up. That is, the graph has a minimum value. Since the minimum of the quadratic equation, $y=a(x-h)^2+k,$ is given by $k,$ then the minimum of the equation above is $2.$ Hence, $y=3x^2-6x+5$ cannot be less than $1.$

the minimum is $2,$ so $3x^2-6x+5$ cannot be less than $1$

$bold{Concept:}$ For a general quadratic function of the form

$f(x)=ax^2+bx+c$

The value of $x$ which maximizes or minimizes $f(x)$ is

$x=-dfrac{b}{2a}$.

The maximum or minimum value of $f(x)$ is the value of $fleft(-dfrac{b}{2a}right)$.

$bold{Solution:}$

a) From the given equation, $P(x)=-5x^2+400x-2550$

$a=-5$ , $b=400$ , $c=-2550$

The value of $x$ that maximizes the profit

$x=-dfrac{b}{2a}=-dfrac{400}{2(-5)}=40$

The maximum profit is the value of $P(x)$ when $x=40$

$P(40)=-5(40)^2+400(50)-2550=5450$
Since $P(x)$ has units of thousand dollars, the maximum profit is $$5,450,000$

b) The amount spent on advertising $x$ was obtained in part (a) as $x=40$.

Since the unit of $x$ is thousand dollars, this amount is $$40,000$

c) To get a profit of atleast $$4,000,000$

we must solve for $x$ such that $P(x)geq 4000$

$-5x^2+400x-2550geq 4000$

$-5x^2+400x-6550geq 0$

$-5(x^2-80x+1310)geq0$

If you multiply/divide both sides by negative number, the inequality symbol is reversed.

$x^2-80x+1310leq 0$

We shall solve the boundaries using quadratic formula

$a=1$, $b=-80$, $c=1310$

$x= dfrac{-bpmsqrt{b^2-4ac}}{2a}$

$x= dfrac{-(-80)pmsqrt{(-80)^2-4(1)(1310)}}{2(1)} =40 pm sqrt{290}$

$x=57.03$ or $x=22.97$

Since $a=1$, the quadratic inequality opens upward. Thus,

$x^2-80x+1310leq 0$ if $x$ is from $x=22.97$ to $x=57.03$.

Therefore, to make sure that you get the desired profit, you must spend at least $$22,971$ but not more than $$57,029$.

a) $$5,450,000$

b) Spend $$40,000$ for maximum profit.

c) Spend at least $$22,971$ but not more than $$57,029$

Sketch the situation as follows. Let $w$ be the width of the rectangular infield area and $h$ is the height.

It follows that $h$ is the diameter of the circle. We need to find the expression for the perimeter of running track (shown in red). Remember that the circumference of a circle is its diameter multiplied by $pi$. If we combine two half-circles, this is just a whole circle with diameter equal to $h$

Perimeter = $2w+picdot h$

The track coach wants this perimeter to be equal to $500$ (since two lapses should be 1000) while the football coach wants to maximize the area of the rectangular infield which is $hcdot w$

Perimeter = $2w+pi cdot h = 1000implies h=dfrac{500-2w}{pi}$

Area of rectangular field = $hw=left(dfrac{500-2w}{pi}right)(w)$

Area = $A(w)=dfrac{1}{pi}(500w-2w^2)$

We shall write the area in the form $f(x)=ax^2+bx+c$

$A(w)=-dfrac{2}{pi}w^2+dfrac{500}{pi}w$

$a=-dfrac{2}{pi}$ , $b=dfrac{500}{pi}$ , $c=0$

The value of $w$ that will maximize the area is $w=-dfrac{b}{2a}$

$w=dfrac{-b}{2a}=-dfrac{500/pi}{2(2/pi)}=125$ meters

$h=dfrac{500-2w}{pi}=dfrac{500-2(125)}{pi}=dfrac{250}{pi}$ meters

Both coaches can be satisfied when $w=125$ m and $h= dfrac{250}{pi}$ m.

Yes. Both coaches can be satisfied when $w=125$ m and $h= dfrac{250}{pi}$ m.

Method 1: The quadratic function is in the form $y=ax^2+bx+c$, so I can find the minimum value by solving for the vertex $(h,k)$ using the formula

$h=-dfrac{b}{2a}$

$k=fleft(dfrac{-b}{2a}right)$

The minimum value is $k$.

In this case, $a=3$, $b=-7$ , $c=2$

$h=-dfrac{b}{2a}=-dfrac{-7}{2(3)}=dfrac{7}{6}$

minimum value: $k=fleft(dfrac{7}{6}right)=3left(dfrac{7}{6}right)^2-7left(dfrac{7}{6}right)+2=-dfrac{25}{12}$

Method 2: Find the zeroes of the function using factoring. The minimum value is the value of $f(x)$ that corresponds to the middle of the two zeroes (axis of symmetry).

$f(x)=3x^2-7x+2$

$f(x)=(3x-1)(x-2)$

$3x-1=0implies x=dfrac{1}{3}$

$x-2=0implies x=2$

The axis of symmetry is therefore

$h=dfrac{1/3+2}{2}=dfrac{7}{6}$

To obtain the minimum value, find $fleft(dfrac{7}{6}right)$

$k=fleft(dfrac{7}{6}right)=3left(dfrac{7}{6}right)^2-7left(dfrac{7}{6}right)+2=-dfrac{25}{12}$

Method 3: You can solve for the zeroes using quadratic formula instead of factoring. You should get the same answer.

Method 4: You can use graphing calculator and find the vertex $(h,k)$. The minimum value is the value of $k$.

Method 5: Use completing the square to rewrite in the form $y=a(x-h)^2+k$.

$f(x)=3x^2-7x+2$

$f(x)=3left(x^2-dfrac{7}{3}xright)+2$

$f(x)=3left[x^2-dfrac{7}{3}x+left(dfrac{7/3}{2}right)^2right]+2-3left(dfrac{7/3}{2}right)^2$

$f(x)=3left[x-left(dfrac{7}{6}right)^2right]-dfrac{25}{12}$

The minimum value is $-dfrac{25}{12}$

I would choose the first method because the value of $a$, $b$, $c$ are readily available.

Method 1: Use the formula for the vertex $left(-dfrac{b}{2a},fleft(dfrac{-b}{2a}right)right)$

Method 2: Find the zeroes using factoring, the value of $x$ halfway to the two zeroes is the axis of symmetry. The minimum value is the value of the function at the axis of symmetry.

Method 3: Use quadratic formula instead of factoring to find the zeroes.

Method 4: Use your graphing calculator to find the minimum value.

Method 5: Use completing the square and write in the form $y=a(x-h)^2+k$

$bold{Concept:}$ For a general quadratic function of the form

$f(x)=ax^2+bx+c$

The value of $x$ which maximizes or minimizes $f(x)$ is

$x=-dfrac{b}{2a}$.

The maximum or minimum value of $f(x)$ is the value of $fleft(-dfrac{b}{2a}right)$.

$bold{Solution:}$

In this case, $a=-4.9$, $b=v_0$ , $c=h_0$

The value of $t$ that corresponds to the maximum height is

$t=-dfrac{b}{2a}=-dfrac{v_0}{2(-4.9)}=dfrac{v_0}{9.8}$ seconds

$t=dfrac{v_0}{9.8}$ seconds

$bold{Concept:}$ For a general quadratic function of the form

$f(x)=ax^2+bx+c$

The value of $x$ which maximizes or minimizes $f(x)$ is

$x=-dfrac{b}{2a}$.

The maximum or minimum value of $f(x)$ is the value of $fleft(-dfrac{b}{2a}right)$.

$bold{Solution:}$

For every $$1$ increase in the price, 30 fewer students attend the dance.

Let $x$ be the increase in price.

ticket price = $8+x$

attending students = $300-30x$

The revenue is the product of ticket price and number of attending students

$R(x)=(8+x)(300-30x)$

We can solve for the zeroes since it is in factored form

$x+8=0implies x=-8$ , $300-30x=0implies x=10$

The axis of symmetry is $x=dfrac{-8+10}{2}=1$

Therefore, the ticket price must be sold at

$$
8+x=8+1=$9
$$

$$
$9
$$