begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|}
hline
$bold{Form}$ & $bold{Family; of; Parabola}$ \ hline
begin{tabular}[c]{@{}l@{}}vertex form\ $f(x)=a(x-h)^2+k$end{tabular} & begin{tabular}[c]{@{}l@{}}if $a$ is varied\ $implies$ families with the same vertex and\ axis of symmetryend{tabular} \ hline
begin{tabular}[c]{@{}l@{}}factored form\ $f(x)=a(x-r)(x-s)$end{tabular} & begin{tabular}[c]{@{}l@{}}if $a$ is varied\ $implies$ families with the same $x$-intercepts\ and axis of symmetryend{tabular} \ hline
begin{tabular}[c]{@{}l@{}}standard form\ $f(x)=ax^2+bx+c$end{tabular} & begin{tabular}[c]{@{}l@{}}if $a$ and $b$ are varied\ $implies$ families with the same $y$-interceptend{tabular} \ hline
end{tabular}
end{table}

This is factored form with varying $a$, thus the family will have the same $x$-intercepts and axis of symmetry

$x$-intercepts and axis of symmetry

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|}
hline
$bold{Form}$ & $bold{Family; of; Parabola}$ \ hline
begin{tabular}[c]{@{}l@{}}vertex form\ $f(x)=a(x-h)^2+k$end{tabular} & begin{tabular}[c]{@{}l@{}}if $a$ is varied\ $implies$ families with the same vertex and\ axis of symmetryend{tabular} \ hline
begin{tabular}[c]{@{}l@{}}factored form\ $f(x)=a(x-r)(x-s)$end{tabular} & begin{tabular}[c]{@{}l@{}}if $a$ is varied\ $implies$ families with the same $x$-intercepts\ and axis of symmetryend{tabular} \ hline
begin{tabular}[c]{@{}l@{}}standard form\ $f(x)=ax^2+bx+c$end{tabular} & begin{tabular}[c]{@{}l@{}}if $a$ and $b$ are varied\ $implies$ families with the same $y$-interceptend{tabular} \ hline
end{tabular}
end{table}

This is vertex form with varying $a$ so they will have the same vertex and axis of symmetry.

They are different in the direction of opening (since $a$ has different sign) and extent of vertical stretching.

same vertex$(2,-4)$ and axis of symmetry.

different in direction of opening and extent of vertical stretching.

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|}
hline
$bold{Form}$ & $bold{Family; of; Parabola}$ \ hline
begin{tabular}[c]{@{}l@{}}vertex form\ $f(x)=a(x-h)^2+k$end{tabular} & begin{tabular}[c]{@{}l@{}}if $a$ is varied\ $implies$ families with the same vertex and\ axis of symmetryend{tabular} \ hline
begin{tabular}[c]{@{}l@{}}factored form\ $f(x)=a(x-r)(x-s)$end{tabular} & begin{tabular}[c]{@{}l@{}}if $a$ is varied\ $implies$ families with the same $x$-intercepts\ and axis of symmetryend{tabular} \ hline
begin{tabular}[c]{@{}l@{}}standard form\ $f(x)=ax^2+bx+c$end{tabular} & begin{tabular}[c]{@{}l@{}}if $a$ and $b$ are varied\ $implies$ families with the same $y$-interceptend{tabular} \ hline
end{tabular}
end{table}

This is in standard form with varying $a$ and $b$, thus what they will have in common is the $y$-intercept which is at $(0,7)$.

$y$-intercept$(0,7)$

If $r$ and $s$ are $x$-intercepts of the parabola, then the equations is

$y=a(x-r)(x-s)$

If the parabola passes through a given point, you can solve for $a$ by substituting the values of the coordinates.

For instance, if it passes through $(u,v)$

$a=dfrac{y}{(x-r)(x-s)}=dfrac{v}{(u-r)(u-s)}$

Then use the value of $a$ to write the final equation.

a)

$y=a(x+4)(x-3)$

Since it passes through $(2,7)$

$7=a(2+4)(2-3)$

$a=dfrac{7}{(6)(-1)}=-dfrac{7}{6}$

Thus, the equation is

$$
y=-dfrac{7}{6}(x+4)(x-3)
$$

b)

$y=a(x)(x-8)$

Since it passes through $(-3,-6)$

$-6=a(-3)(-3-8)$

$a=dfrac{-6}{(-3)(-11)}$

$a=-dfrac{6}{33}=-dfrac{2}{11}$

$y=-dfrac{2}{11}x(x-8)$

c)

$y=a(x-sqrt{7})(x+sqrt{7})$

Since it passes through $(-5,3)$

$3=a(-5-sqrt{7})(-5+sqrt{7})$

$a=dfrac{3}{(-5-sqrt{7})(-5+sqrt{7})}$

$a=dfrac{3}{(-5)^2-(sqrt{7})^2}$

$a=dfrac{3}{25-7}$

$a=dfrac{3}{18}=dfrac{1}{6}$

Therefore, the equation is

$y=dfrac{1}{6}(x+sqrt{7})(x-sqrt{7)}$

$y=dfrac{1}{6}(x^2-7)$

$y=a(x-1+sqrt{2})(x+1-sqrt{2})$

$y=a[x^2-x(1+sqrt{2})-x(1-sqrt{2})+(1-sqrt{2})(1+sqrt{2})]$

$y=a[x^2-x-sqrt{2}-x+sqrt{2}+(1+sqrt{2}-sqrt{2}-sqrt{4})]$

$y=a[x^2-2x+(1-2)]$

$f(x)=a(x^2-2x-1)$

Since it passes through $(2,4)$

$4=a(2^2-2cdot 2-1)$

$a=dfrac{4}{4-4-1}$

$a=-4$

Therefore, the equation is

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

a) $f(x)=-dfrac{7}{6}(x+4)(x-3)$

b) $f(x)=-dfrac{2}{11}x(x-8)$

c) $f(x)=dfrac{1}{6}(x^2-7)$

d) $f(x)=-4(x^2-2x-1)$

If a parabola has a vertex$(h,k)$ then the equation is

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

If it passes through the point $(u,v)$, then you can evaluate $a$ by
substituting the coordinates to the equation

$v=a(u-h)^2+k$

$$
a=dfrac{v-k}{(u-h)^2}
$$

a) $y=a(x+2)^2+5$

Since it passes through $(4,-8)$

$-8=a(4+2)^2+5$

$a=dfrac{-8-5}{(4+2)^2}=-dfrac{13}{36}$

Therefore, the equation is

$$
y=-dfrac{13}{36}(x+2)^2+5
$$

b) $y=a(x-1)^2+6$

Since it passes through $(0,-7)$

$-7=a(0-1)^2+6$

$a=dfrac{-7-6}{(-1)^2}=-13$

Therefore, the equation is

$$
y=-13(x-1)^2+6
$$

c) $y=a(x-4)^2-5$

Since it passes through $(-1,-3)$

$-3=a(-1-4)^2-5$

$a=dfrac{-3+5}{(-1-4)^2}=dfrac{2}{25}$

Therefore, the equation is

$y=dfrac{2}{25}(x-4)^2-5$

d) $y=a(x-4)^2$

Since it passes through $(11,8)$

$8=a(11-4)^2$

$a=dfrac{8}{7^2}=dfrac{8}{49}$

Therefore, the equation is

$$
y=dfrac{8}{49}(x-4)^2
$$

a) $y=-dfrac{13}{36}(x+2)^2+5$

b) $y=-13(x-1)^2+6$

c) $y=dfrac{2}{25}(x-4)^2-5$

d) $y=dfrac{8}{49}(x-4)^2$

Given that $f(2)=3$ for $f(x)=ax^2-6x-7$

Substitute $x=2$ and $f(x)=3$, then solve for $a$

$3=a(2)^2-6(2)-7$

$a=dfrac{7+6(2)+3}{2^2}=5.5$

Therefore, the equation is

$$
f(x)=5.5x^2-6x-7
$$

$$
f(x)=5.5x^2-6x-7
$$

a) Since it is given in factored form, we can readily obtain the zeros at $(2,0)$ and $(-6,0)$. The midpoint between these points must be the correspond to the axis of symmetry which is $h=dfrac{2-6}{2}=-2$. To find the vertex$(h,k)$, we can use the fact that $k$ is the value of $f(x)$ at the axis of symmetry, that is, $k=f(h)$

$f(-2)=(-2-2)(-2+6)=-16$

Thus, the vertex is $(-2,-16)$. As expected, this graph must open upward.

With this information, we can sketch the graph as follows.

b) Since it is reflected in the $x$-axis, the $x$-intercepts are still the same but the vertex will be $(-2,16)$

c) The $x$-intercepts will be the same as in part(a) but there will be vertical stretching by a factor of 3, so the distance of the vertex to the $x$-axis will be 3 times as much.

a) $x$-intercepts($-6,0$) and $(2,0)$, vertex$(-2,-16)$

b) $x$-intercepts($-6,0)$ and $(2,0)$, vertex$(-2,16)$

c) $x$-intercepts($-6,$0) and $(2,0)$, vertex$(-2,-48)$

If $r$ and $s$ are $x$-intercepts of the parabola, then the equations is

$y=a(x-r)(x-s)$

If the parabola passes through a given point, you can solve for $a$ by substituting the values of the coordinates.

For instance, if it passes through $(u,v)$

$a=dfrac{y}{(x-r)(x-s)}=dfrac{v}{(u-r)(u-s)}$

Then use the value of $a$ to write the final equation.

Since the zeros are $4$ and $-4$,

$y=a(x-4)(x+4)$

Since it passes through $(3,6)$

$6=a(3-4)(3+4)$

$6=a(-1)(7)$

$a=-dfrac{6}{7}$

Therefore, the equation is

$y=-dfrac{6}{7}(x-4)(x+4)$

$$
y=-dfrac{6}{7}(x^2-16)
$$

$$
y=-dfrac{6}{7}(x^2-16)
$$

If $r$ and $s$ are $x$-intercepts of the parabola, then the equations is

$y=a(x-r)(x-s)$

If the parabola passes through a given point, you can solve for $a$ by substituting the values of the coordinates.

For instance, if it passes through $(u,v)$

$v=a(u-r)(u-s)$

$a=dfrac{v}{(u-r)(u-s)}$

Then use the value of $a$ to write the final equation.

In this case,

$y=a[x-(2+sqrt{3})][x-(2-sqrt{3})]$

$y=a[x^2-(2-sqrt{3})x-(2+sqrt{3})x+(2+sqrt{3})(2-sqrt{3})]$

$y=a[x^2-4x+(2^2-(sqrt{3})^2)]$

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

Now, substitute $x=-4$ and $y=5$ and solve for $a$

$5=a[(-4)^2-4(-4)+1]$

$5=a(33)$

$a=dfrac{5}{33}$

Therefore, the equation is

$y=dfrac{5}{33}(x^2-4x+1)$

$$
y=dfrac{5}{33}(x^2-4x+1)
$$

Let the left-bottom edge of the arch be the origin$(0,0)$. Since the tunnel is 12 m wide, the other edge must have coordinates $(12,0)$. With these zeros, the equation is

$y=a(x-0)(x-12)$

$y=ax(x-12)$

To evaluate $x$, we will use the fact that the parabola passes through $(4,6)$

$6=a(4)(4-12)$

$a=dfrac{6}{4(-8)}=-dfrac{3}{16}$

Therefore, the equation describing the parabolic arch is

$y=-dfrac{3}{16}x(x-12)$

To determine whether a truck 5 m tall and 3.5 wide can pass through, calculate the distance between two points for which $y=5$. If this distance is greater than $3.5$, then the truck can pass through.

$5=-dfrac{3}{16}(x)(x-12)$

$80=-3x(x-12)$

$80=-3x^2+36x$

$0=-3x^2+36x-80$

We will use quadratic formula

$a=-3$ , $b=36$ , $c=-80$

$x=dfrac{-36pmsqrt{36^2-4(-3)(-80)}}{2(-3)}$

$x=dfrac{-36pmsqrt{1296-960}}{-6}$

$x=dfrac{-36pmsqrt{336}}{-6}$

$x=dfrac{-36pm 18.33}{-6}$

$x=2.94$ and $x=9.05$

The difference between these zeros is $9.05-2.94=6.11$

Since the width is greater than $5$ m, the truck can pass through.

$f(x)=-dfrac{3}{16}x(x-12)$

Yes, since at a height of 5 m, the width of the tunnel is $6.11$ m.

Vertex is at $(3,56)$ so we can write the vertex form as

$f(x)=a(t-3)^2+56$

We can choose any point on the table to evaluate $a$. This time, we will choose $(0,11)$

$11=a(0-3)^2+56$

$11=9a+56$

$9a=-45$

$a=-5$

Therefore, the equation describing the data is

$$
y=-5(t-3)^2+56
$$

a) see scatter plot inside

b) see graph inside

c) $y=-5(t-3)^2+56$

Vertex is at $(1,5)$ so we can write the vertex form as

$f(x)=a(t-1)^2+5$

We can choose any point on the table to evaluate $a$. This time, we will choose $(0,0)$

$0=a(0-1)^2+5$

$0=a+5$

$a=-5$

Therefore, the equation describing the data is

$y=-5(t-1)^2+5$

$y=-5(t^2-2t+1)+5$

$y=5t^2+10t-5+5$

$$
y=5t^2+10t
$$

a) see scatter plot inside

b) see graph inside

c) $y=5t^2+10t$

b) The vertex is estimated to be at $(1.35,442)$

Vertex is at $(1.35,442)$ so we can write the vertex form as

$f(x)=a(t-1.35)^2+442$

We can choose any point on the table to evaluate $a$. This time, we will choose $(0.60,198)$

$35=a(0.30-1.35)^2+442$

$a=dfrac{35-442}{(0.3-1.35)^2}=-433.78$

Therefore, the equation describing the data points is

$y=-433.78(x-1.35)^2+442$

Note that the answer at the back of your book is $f(x)=343(x-1.35)^2+442$ which is impossible since it should open downwards.

It’s also worth noting that in this case, we have used the vertex and an arbitrary point on the parabola to find the equation. In practice, this problem must be solved by non-linear regression to account for all data points (method of least-squares) which is beyond the scope of this text.

If we use non-linear regression, we obtain

$y=-343.92(x-1.40368)^2+434.391$

shown in green curve.

Our solution using only one point is shown in red.

a) see scatter plot inside

b) approximately $(1.35,442)$

c) $f(x)=-433.78(x-1.35)^2+442$

If $(h,k)$ is the vertex of a parabola, then the vertex form can be written as

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

From the graph, the vertex is $(-2,3)$ so the vertex form is

$y=a(x+2)^2+3$

Since $(-4,-9)$ is a also point on the parabola,we can evaluate $a$ by substituting this point to the vertex form

$-9=a(-4+2)^2+3$

$-9=a(4)+3$

$a=dfrac{-9-3}{4}=-3$

Therefore, the equation of the graph is

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

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

$bold{Definition:}$

A group of parabolas having a common characteristic such as vertex, zeros, and $y$-intercept.

$bold{Characteristics:}$

Family of parabolas may share zeros, a vertex, or a $y$-intercept.

In $bold{vertex; form}$ $f(x)=a(x-h)^2+k$

a family of parabola with the same vertex can be obtained by varying $a$.

In $bold{factored; form}$ $y=a(x-r)(x-s)$

a family with the same zeros can be obtained by varying $a$

In $bold{standard; form}$, $y=ax^2+bx+c$, a family of parabola with the same $y$-intercept can be obtained by varying $a$ and/or $b$

$bold{Examples:}$

vertex form (same vertex):

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

$g(x)=-9(x-1)^2+3$

factored form (same zeros):

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

$q(x)=9(x-2)(x+3)$

$f(x)=23.5(x+3)(x-2)$

standard form (same $y$-intercept):

$f(x)=2x^2+3x+4$

$$
g(x)=-5x^2+78x+4
$$

$bold{Nontext{-}examples:}$

vertex form:

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

$g(x)=2(x+1)^2+3$

factored form:

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

$p(x)=6(x+3)(x-1)$

$p(x)=6(x+3)(x+2)$

standard form:

$f(x)=2x^2+3x+4$

$$
f(x)=3x^2+3x+5
$$

If $(r,0)$ and $(s,0)$ are zeros of a quadratic function, then the factored form can be written as

$y=a(x-r)(x-2)$

Let the point $(0,0)$ be one edge of the bridge. Since the tunnel is 40 m wide, the other zero must be at $(40,0)$

Thus, we can write the equation in factored form as

$y=a(x-0)(x-40)$

$y=ax(x-40)$

To find $a$ we shall substitute any point on the parabola to the equation.

Since the height is 8 m at a point 5 m from the edge, then $(5,8)$ must be a point on the parabola.

$8=a(5)(5-40)$

$8=5a(-35)$

$8=-175a$

$a=-dfrac{8}{175}$

Therefore, the equation describing the bridge is

$y=-dfrac{8}{175}x(x-40)$

We want to know the height of the bridge when $x=12$

$y=-dfrac{8}{175}(12)(12-40)$

$h=15.36$ m

Therefore, the height of the bridge 12 m from the edge is $15.36$ m

$15.36$ m

Just substitute $x=3$ and $f(x)=6$ to the cubic function and solve for $a$

$f(x)=a(x+3)(x-1)(x-5)$

$6=a(3+3)(3-1)(3-5)$

$6=a(6)(2)(-2)$

$a=dfrac{6}{6(2)(-2)}=-dfrac{1}{4}$

Therefore, the equation is

$$
f(x)=-dfrac{1}{4}(x+3)(x-1)(x+5)
$$

$$
f(x)=-dfrac{1}{4}(x+3)(x-1)(x+5)
$$