b) Since the second differences are constant, it is quadratic.

d) Since the second differences are constant, it is quadratic.

$a>0implies$ opens up

$a<0implies$ opens down

b) $f(x)=-2(x-3)(x+1)implies a=-3implies$ opens down

c) $f(x)=-(x+5)^2implies a=-1implies$ opens down

d) $f(x)=dfrac{2}{3}x^2-2x-1implies a=dfrac{2}{3}implies$ opens up

Here, $f(x)$ is already in factored form

a) $(x-2)=0implies x=2$

$(x+6)=0implies x=-6$

b) $a=-3$ so it should open downwards

c) We can obtain the axis of symmetry by rewriting in the form

$f(x)=a(x-h)^2+k$

The axis of symmetry is $x=h$

$y=-3(x-2)(x+6)$

$y=-3(x^2+6x-2x-12)$

$y=-3(x^2+4x-12)$

Complete the square by adding and subtracting 4

$y=-3[(x^2+4x+4)-12-4]$

$y=-3[(x+2)^2-16]$

$y=-3(x+2)^2+48$

Thus, the axis of symmetry is $x=-2$

b) opens downward

c) $x=-2$

From the graph, vertex($-2,3$)

b) The axis of symmetry is the line that divides the parabola into two equal parts

From the graph, $x=-2$

c) The domain is the set of all possible values of $x$ which is all real numbers for all quadratic functions.

domain: $bold{R}$

The range is the set of all possible values of $y$.

Since the graph opens downward,

range: ${ yin bold{R};|;yleq 3}$

b) $x=-2$

c) domain: $bold{R}$

range: ${ yin bold{R};|;yleq 3}$

$bold{axis; of;symmetry}$ is the line that divides the parabola into two equal parts. If the vertex is $(h,k)$, the axis of symmetry is $x=h$

a)

direction of opening: upward

vertex: $(0,-3)$

axis of symmetry: $x=0$

direction of opening: downward

vertex: $(-3,-4)$

axis of symmetry: $x=-3$

direction of opening: upward

vertex$(1,-18)$

axis of symmetry: $x=1$

direction of opening: downwards

vertex$(0,4)$

axis of symmetry: $x=0$

b) opens down, vertex$(-3,-4)$, $x=-3$

c) opens up, vertex$(1,-18)$, $x=1$

d) opens down, vertex$(0,4)$, $x=0$

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

$f(x)=-3(x^2-2x+1)+6$

$f(x)=-3x^2+6x-3+6$

$f(x)=-3x^2+6x+3$

The $y$-intercept is the point on the graph that crosses $y$-axis.
This corresponds to the point at $x=0$

$f(0)=-3(0)^2+6(0)+3=3$

y-intercept: $(0,3)$

$f(x)=4(x^2-7x-3x-21)$

$f(x)=4(x^2-11x-21)$

$f(x)=4x^2-44x-84$

$f(0)=4(0)^2-44(0)-84=-84$

$y$-intercept: $(0,-84)$

b) $f(x)=4x^2-44x-84$ ; $(0,-84)$

a) The parabola opens downward.

b) In this case, vertex is the maximum point of the graph which is vertex$(-1,8)$

c) The values of $x$-intercepts are the points where $y=0$

$xtext{-intercepts}$ : $(-3,0)$ , $(1,0)$

d) The domain of all quadratic function is $bold{R}$

domain: ${xin bold{R}}$

Since the graph opens downward, the range is all real number less than the maximum value at the vertex.

range: ${ y in bold{R};|;yleq 8}$

e) Since the graph opens downward, the second differences is negative.

f) Use the vertex form

$y=a(x-h)^2+k$

$y=a(x+1)^2+8$

To find $a$, substitute any point on the graph. Here we will choose $(1,0)$

$0=a(1+1)^2+8implies 4a=-8implies a=-2$

$$
y=-2(x+1)^2+8
$$

b) vertex$(-1,8)$

c) $(-3,0)$, $(1,0)$

d) domain: ${ xin bold{R}}$ , range $={ yin bold{R};|;yleq 8}$

e) negative; opens downwards

f) $f(x)=-2(x+1)^2+8$

a) The parabola opens upwards.

b) In this case, vertex is the minimum point of the graph which is vertex$(1,-3)$

c) The axis of symmetry is $x=himplies x=1$

d) The domain of all quadratic function is $bold{R}$

domain: ${xin bold{R}}$

Since the graph opens upwards, the range is all real number greater than the minimum value at the vertex.

range: ${ y in bold{R};|;ygeq -3}$

e) Since the graph opens downward, the second differences is positive.

b) vertex$(1,-3)$

c) $x=1$

d) domain: ${ xin bold{R}}$ , range $={ yin bold{R};|;ygeq -3}$

e) positive; because it opens upwards

b) $x=-7$

c) $x=12$

d) $x=-2$

e) $x=-dfrac{3}{2}$

f) $x=-dfrac{5}{16}$

b) We could predict the sign from the direction of opening of the graph. Since it opens downwards, the second differences must be negative.\\

$y=a(x+1)^2+4$

Find the value of $a$ using any point on the graph. We’ll choose $(1,0)$

$0=a(1+1)^2+4$

$0=a(2)^2+4$

$0=4a+4implies a=dfrac{4}{-4}=-1$

The equation of the graph is

$$
y=-(x+1)^2+4
$$

$x=3$ is the location of $x$-intercept $implies f(3)=0$,
that is, graph passes through (3,0)

$y=32$ is the maximum value $implies$ $k=32$, opens down

vertex $(-1,32)$

From the vertex form, $y=a(x-h)^2+k$,

$y=a(x+1)^2+32$

we can substitute the $(3,0)$ to find $a$

$0=a(3+1)^2+32$

$16a=-32implies a=-2$

Thus, the equation is

$y=-2(x+1)^2+32$

The $y$-intercept is the point where the graph cross the $y$-axis, that is, when $x=0$

$y=-2(0+1)^2+32$

$$
y=30
$$

$f(x)=2x^2-4x=2x(x-2)$ is a quadratic function in factored form whose zeros are at $x=0$ and $x=2$. The axis of symmetry lies midway between two zeros which is at $x=dfrac{0+2}{2}implies x=1$. We can find the vertex by solving for $f(x)$ at $x=1$ which is $f(1)=-2$, thus its vertex is $(1,-2)$. This function has $a>0$, so it opens upward.

$g(x)=-(x-1)^2+2$ is also a quadratic function in the vertex form $y=a(x-h)^2+k$ where $a=-1$ (opens downward), $h=1$ (axis of symmetry) and vertex at $(1,2)$.

Differences: $f(x)$ opens upward while $g(x)$ opens downward. $f(x)$ has vertex at $(1,-2)$ while $g(x)$ has vertex at $(1,2)$

Vertex form: $P(x)=a(x-h)^2+k$ $implies$ vertex$(h,k)$

Factored form: $P(x)=a(x-r)(x-s)$ $implies$ $h=dfrac{1}{2}(r+s)$ , $k=P(h)$

For computer sales,

$P(x)=-2(x-3)^2+50$

Thus, the maximum profit for computer sales is $$50,000$

For stereo systems,

$P(x)=-(x-2)(x-7)$

$h=dfrac{2+7}{2}=4.5$

$k=P(4.5)=-(4.5-2)(4.5-7)=6.25$

Thus, the maximum profit for stereo systems is $$6,250$

Therefore, the maximum profit of the company is

$$
50,000+6,250=$56,250
$$

$y=a(x+60)(x-35)$

Now, we shall find the value of $a$ by substituting the point $(0,20)$

$20=a(0+60)(0-35)implies a=dfrac{20}{(60)(-35)}=-dfrac{1}{105}$

Therefore, the equation of the parabola is

$y=-dfrac{1}{105}(x+60)(x-35)$

Note that the answer at the back of your back is wrong and it has been corrected in the published errata.