$bold{Concept:}$ The vertex of quadratic function of the form

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

is vertex$(h,k)$

If $a>0$, the graph opens upward while if $a0$, there are two real zeros.

If $b^2-4ac=0$, there is only one real zero.

If $b^2-4ac<0$, there are no real zeros.

$bold{Solution:}$

a) $f(x)=3(x-0)^2-5$

$a=3$, $b=0$ , $c=-5$

By inspection, the vertex is $(0,-5)$

Since $a>0$, the graph opens upward.

$b^2-4ac=0^2-4(3)(-5)=60>0$

There are two real zeros.

b) $f(x)=-4(x-0)^2+7$

$a=-4$ , $b=0$ , $c=7$

By inspection, the vertex is $(0,7)$

Since $a0$

There are two real zeros.

c) $f(x)=5(x-0)^2+3$

$a=5$, $b=0$ , $c=3$

By inspection, the vertex is $(0,3)$

Since $a>0$, the graph opens upward.

$b^2-4ac=0^2-4(5)(3)=-60<0$

There are no real zeros.

d) $f(x)=3(x+2)^2+0$

By inspection the vertex is $(-2,0)$

Since $a>0$, the graph opens upward.

We can obtain the zeros as

$3(x+2)^2=0$

$(x+2)=pmsqrt{0/3}$

$x=-2$

There is only one zero.

e) $f(x)=-4(x+3)^2-5$

By inspection the vertex is $(-3,-5)$

$f(x)=-4(x+3)^2-5$

$f(x)=-4(x^2+6x+9)-5$

$f(x)=-4x^2+24x-39$

Since $a<0$, the graph opens upward.

$b^2-4ac=24^2-4(-4)(-39)=-48<0$

No real zeros.

f) $f(x)=0.5(x-4)^2-2$

By inspection, the vertex is $(4,-2)$

$f(x)=0.5(x^2-8x+16)-2$

$f(x)=0.5x^2-4x+6$

Since $a>0$, the graph opens upward.

$b^2-4ac=(-4)^2-4(0.5)(6)=4>0$

There are two real zeros.

a) vertex$(0,-5)$, up, 2 zeros

b) vertex$(0,7)$, down, 2 zeros

c) vertex$(0,3)$, up, no zeros

d) vertex$(-2,0)$, up, 1 zero

e) vertex$(-3,-5)$, down, no zeros

f) vertex$(4,-2)$, up, 2 zeros

We shall solve the following using factoring to determine the number of zeros.

a) $f(x)=x^2-6x-16=(x-8)(x+2)$

$x=8$ or $-$2 $implies 2$ zeros

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

$x=0$ or 3 $implies$ 2 zeros

c) $f(x)=4x^2-1=(2x)^2-1^2=(2x+1)(2x-1)$

$x=-dfrac{1}{2}$ or $dfrac{1}{2}implies$ 2 zeros

d) $f(x)=9x^2+6x+1=(3x)^2+2(3x)(1)+1^2=(3x+1)^2$

$x=-dfrac{1}{3}implies$ one zero

$bold{Concept:}$ To determine the number of zeros, use the following guideline

$b^2-4ac>0implies$ 2 zeros

$b^2-4ac=0implies$ 1 zero

$b^2-4ac<0implies$ no zeros

a) $a=2$, $b=-6$, $c=-7$

$b^2-4ac=(-6)^2-4(2)(-7)=92implies$ 2 zeros

b) $a=3$ , $b=2$, $c=7$

$b^2-4ac=2^2-4(3)(7)=-80implies$ no zeros

c)$a=1$ , $b=8$ , $c=16$

$b^2-4ac=8^2-4(1)(16)=0implies$ 1 zero

d) $a=9$ , $b=-14.4$, $c=5.76$

$b^2-4ac=(-14.4)^2-4(9)(5.76)=0implies$ 1 zero

We have to use different methods for determining number of zeros in each cases.

Method 1: Determine $b^2-4ac$

Method 2: Solve by factoring

Method 3: Solve by quadratic formula

Method 4: Graphing

a) We will use graphing.

In this case, $a=-3<0$, so it opens downward and vertex is at $(2,4)$.

From the graph, there are two zeros.

b) We will use factoring.

$f(x)=5(x-3)(x+4)$

$x=3$ or $-4$

There are two zeros.

c) We will use quadratic formula

$a=4$, $b=-2$, $c=0$

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

$x=dfrac{2pmsqrt{2^2-4(4)(0)}}{2(4)}$

$x=dfrac{2pm 2}{8}$

$x=0$ or $x=dfrac{1}{2}$

There are two zeros.

d) We will determine $b^2-4ac$

$a=3$ , $b=-1$ , $c=5$

$b^2-4ac=(-1)^2-4(3)(5)=-59<0$

There are no zeros.

$bold{Concept:}$ Given the profit function, the number of ways that it can break even is the number of zeros. If there are no zeros, it cannot break even.

To determine the number of zeros, use the following guidelines:

$b^2-4ac>0implies$ 2 zeros

$b^2-4ac=0implies$ 1 zero

$b^2-4ac<0implies$ no zeros

a) $a=-2.1$ , $b=9.06$, $c=-5.4$

$b^2-4ac=9.06^2-4(-2.1)(-5.4)=36.7236>0$

Two ways to break-even.

b) $a=-0.3$ , $b=2$ , $c=-7.8$

$b^2-4ac=2^2-4(0.3)(-7.8)=-5.36<0$

The company cannot break-even.

c) $a=-2$, $b=6.4$, $c=-5.124$

$b^2-4ac=6.4^2-4(-2)(-5.12)=0$

One way to break-even.

d) $a=-2.4$ , $b=1$ , $c=-1.2$

$b^2-4ac=1^2-4(-2)(-1.2)=-8.6<0$

The company cannot break-even.

$bold{Concept:}$ We can obtain the number of $x$-intercepts by calculating $b^2-4ac$

$b^2-4ac>0implies$ 2 $x$-intercepts

$b^2-4ac=0implies$ 1 $x$-intercept

$b^2-4ac<0implies$ no $x$-intercept

$bold{Solution:}$ In order to have one $x$-intercept $b^2-4ac$

In this case,

$a=3$ , $b=-4$, $c=k$

$b^2-4ac=(-4)^2-4(3)(k)=0$

$16-12k=0$

$$
k=dfrac{4}{3}
$$

$$
k=dfrac{4}{3}
$$

$bold{Concept:}$ We can obtain the number of $x$-intercepts by calculating $b^2-4ac$

$b^2-4ac>0implies$ 2 zeros

$b^2-4ac=0implies$ 1 zero

$b^2-4ac<0implies$ no zeros

$bold{Solution:}$ In order to have no zeros $b^2-4ac<0$

In this case,

$a=k$ , $b=-4$, $c=k$

$b^2-4ac=(-4)^2-4(k)(k)<0$

$16-4k^2<0$

$-4k^216$

$k^2>4$

$k2$

$k2$

$bold{Concept:}$ We can obtain the number of $x$-intercepts by calculating $b^2-4ac$

$b^2-4ac>0implies$ 2 zeros

$b^2-4ac=0implies$ 1 zero

$b^2-4ac<0implies$ no zeros

$bold{Solution:}$ In this case,

$a=3$ , $b=4$, $c=k$

$b^2-4ac=4^2-4(3)(k)=16-12k$

2 zeros: $16-12k>0implies k<dfrac{4}{3}$

1 zero: $16-12k=0implies k=dfrac{4}{3}$

no zeros: $16-2kdfrac{4}{3}$

2 zeros: $kdfrac{4}{3}$

$bold{Concept:}$ We can obtain the number of $x$-intercepts by calculating $b^2-4ac$

$b^2-4ac>0implies$ 2 zeros

$b^2-4ac=0implies$ 1 zero

$b^2-4ac<0implies$ no zeros

$bold{Solution:}$ If $f(x)$ touches $x$-axis at one point, then $b^2-4ac=0$

$a=1$, $b=-k$, $c=k+8$

$b^2-4ac=(-k)^2-4(1)(k+8)=0$

$k^2-4k-32=0$

$(k-8)(k+4)=0$

$k=8$ or $-$4

$k=8$ or $-4$

$bold{Concept:}$ We can obtain the number of solutions to a quadratic equation by calculating $b^2-4ac$

$b^2-4ac>0implies$ 2 solutions

$b^2-4ac=0implies$ 1 solution

$b^2-4ac<0implies$ no solutions

$bold{Solution:}$ Write the equation in form $ax^2+bx+c=0$

$n^2+25=-8n$

$n^2+8n+25=0$

$a=1$, $b=8$ , $c=25$

$b^2-4ac=8^2-4(1)(25)=-36<0$

It has no solution, thus it is NOT possible that $n^2+25=-8n$

$bold{Concept:}$ We can obtain the number of $x$-intercepts (zeros) by calculating $b^2-4ac$

$b^2-4ac>0implies$ 2 zeros

$b^2-4ac=0implies$ 1 zero

$b^2-4ac<0implies$ no zeros

$bold{Solution:}$

a) We can write the parabola with 2 zeros in factored form such that $a<0$

For example:

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

$f(x)=-(x+6)(x+32)$

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

b) A parabola that has no zeros should have $b^2-4ac0$

For example:

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

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

$$
f(x)=(x+2)^2+12
$$

c) When the vertex is also the zero, there is only 1 zero so $b^2-4ac=0$

Since it opens down, $a<0$

For example

$f(x)=-x^2$

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

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

Answers can vary.

a) $f(x)=-2(x-3)(x+5)$

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

c) $f(x)=-x^2$

The recommended machine is the one that results in the earliest break-even point. The break-even point occurs when the revenue is equal to the cost function. The revenue function is the product of number of units sold and demand function.

Thus, we shall find the value of $x$ such that $xcdot p(x)=C(x)$

Machine A:

$-4x+42.5=4.1x+92.16$

$x(-4x+42.5)=4.1x+92.16$

$-4x^2+42.5x=4.1x+92.16$

$-4x^2+38.4x-92.16=0$

We will use quadratic formula

$a=-4$ , $b=38.4$ , $c=-92.16$

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

$x=dfrac{-38.4pmsqrt{38.4^2-4(-4)(-92.16)}}{2(-4)}$

$x=dfrac{38.4pmsqrt{0}}{8}$

$x=4.8$

Machine B:

$x(-4x+42.5)=17.9x+19.36$

$-4x^2+42.5x=17.9x+19.36$

$4x^2-24.6x+19.36=0$

$a=4$, $b=-24.6$ , $c=19.36$

$x=dfrac{-(-24.6)pmsqrt{(-24.6)^2-4(4)(19.36)}}{2(4)}$

$x=dfrac{24.6pmsqrt{605.16-309.76}}{8}$

$x=dfrac{24.6pmsqrt{295.4}}{8}$

$x=dfrac{24.6pm 17.187}{8}$

$x=0.93$ or $5.2$

The first break-even point is $0.93$

Machine C:

$x(-4x+42.5)=8.8x+55.4$

$-4x^2+42.5x=8.8x+55.4$

$4x^2-33.7x+55.4=0$

$a=4$ , $b=-33.7$ , $c=55.4$

$x=dfrac{-(-33.7)pmsqrt{(-33.7)^2-4(4)(55.4)}}{2(4)}$

$x=dfrac{33.7pmsqrt{1135.69-886.4}}{8}$

$x=dfrac{33.7pmsqrt{249.29}}{8}$

$x=dfrac{33.7-15.7889}{8}$

$x=6.2$ or $x=2.2$

The first break-even point is 2.2.

Machine A: break-even $x=4.8$

Machine B: break-even $x=0.93$

Machine C: break-even $x=2.2$

Choose Machine B since it has the earliest break-even point.

If the vertex$(h,k)$ is at the $x$-axis, there is only one zero.

If the vertex$(h,k)$ is above the $x$-axis and it opens downward, there will be two zeros.

If the vertex$(h,k)$ is below the $x$-axis and it opens upward, there will be two zeros.

If none of the above conditions are true, there would be no zeros.

a) Vertical stretching does not change the location of vertex. The vertex is still on the $x$-axis, so it would have no effect.

b) Horizontal translation will move the vertex along the $x$-axis, thus it would still be on the $x$-axis, and would have no effect.

c) Horizontal compression will not change the location of the vertex. Reflection in the $x$-axis will not change the vertex since it’s on the $x$-axis. Thus, it would have no effect.

d) Vertical translation 3 units down would result in a vertex below the $x$-axis opening upward. Thus, there will be two zeros.

e) Horizontal translation 4 units to the right would result in a vertex on the axis. However, vertical translation 3 units up would result in a vertex above the $x$-axis and the parabola opens upwards. In this case, there will be no zeros.

f) Reflection in the $x$-axis would result in parabola which opens downwards while horizontal translation 1 unit to the left would still result in vertex lying in the $x$-axis. A vertical translation 5 units up will set the vertex above the $x$-axis and the parabola opens downward, thus, there will be 2 zeros.

If we have two functions $y=f(x)$ and $y=g(x)$, then the intersection points of the graph of $f(x)$ and $g(x)$ is obtained by setting $f(x)=g(x)$ and the value of $x$ that satisfies the equation correspond to the intersection points.

In this case,

$f(x)=g(x)$

$x^2-6x+14=-x^2-20x-k$

$2x^2+14x+14+k=0$

The two parabolas must intersect only at one point, thus, we shall find the value of $k$ which results in only one zero.

Note that a quadratic function has only one zero if $b^2-4ac=0$

$a=2$ , $b=14$, $c=14+k$

$b^2-4ac=14^2-4(2)(14+k)=0$

$14+k=dfrac{14^2}{4(2)}$

$k=dfrac{14^2}{4(2)}-14$

$k=10.5$

$k=10.5$

Observe that $f(x)=4-(x-3)(3x+1)$ is a vertical translation 4 units upward of the function $g(x)=-(x-3)(3x+1)$

The parent function $g(x)$ has 2 zeros and opens downward. Vertical translation will not affect the number of zeros, so $f(x)$ should have 2 zeros.

a) The vertex form is $f(x)=a(x-h)^2+k$

If $k=0$, there will be only one zero since the vertex is at $x$-axis.

If $a$ and $k$ have different sign, there will be two zeros.

If $a$ and $k$ have the same sign, there will be no zeros.

b) If the linear factors are equal or multiples of each other, then there will be only 1 zero. If not, there are 2 zeros.

In other words, there will be 1 zero if it can be written in the form
$y=a(x+s)^2$, otherwise, there will be 2 zeros.

c) The standard form is written as $y=ax^2+bx+c$

Calculate $b^2-4ac$, if

$b^2-4ac=0implies$ 1 zero

$b^2-4ac>0implies$ 2 zeros

$b^2-4ac<0implies$ no zeros

a)
If $k=0$, there will be only one zero since the vertex is at $x$-axis.

If $a$ and $k$ have different sign, there will be two zeros.

If $a$ and $k$ have the same sign, there will be no zeros.

b) One zero if it can be written in the form
$y=a(x+s)^2$, otherwise, there will be 2 zeros

c) $b^2-4ac=0implies$ 1 zero

$b^2-4ac>0implies$ 2 zeros

$b^2-4ac<0implies$ no zeros

We shall rewrite $(x^2-1)k=(x-1)^2$ in the form $ax^2+bx+c$

$kx^2-k=x^2-2x+1$

$kx^2-x^2+2x-k-1=0$

$x^2(k-1)+2x-(k+1)=0$

$a=k-1$ , $b=2$, $c=-(k+1)$

Now, we shall calculate $b^2-4ac=0$

If two values of $k$ can satisfy $b^2-4ac=0$, then there will be two solutions

If only one value of $k$ can satisfy $b^2-4ac=0$, then it has only one solution.

$b^2-4ac=2^2-4(k-1)[-(k+1)]=0$

$4+4(k^2-1)=0$

$4(k^2-1)=-4$

$k^2-1=-1$

$k^2=0$

$k=0$

Thus, $(x^2-1)k=(x-1)^2$ has one solution for only

one value of $k$ which is $k=0$

$b^2-4ac=0$ only when $k=0$

We shall find $b^2-4ac$ for the given function.

If $b^2-4ac=0implies$ 1 zero

$b^2-4ac>0implies$ 2 zeros

$b^2-4ac0$ for all values of $k$, thus it will always have 2 zeros.

There will be no value of $k$ that will result in no zeros or one zero.

2 zeros for all values of $k$