a) We can factor this expression

$f(x)=-5(x^2+10x-5)=-5(x^2-2x+1)$

$f(x)=-5(x-1)(x-1)$

The vertex$(h,k)$ of quadratic function is at value the $x$ equal to the average of two zeros.

$h=dfrac{x_1+x_2}{2}=dfrac{1+1}{2}=1$

Substitute this value of $x$

$k=f(h)=-5(1)^2+10(1)-5=0$

Thus, the vertex is $(1,0)$

In this case, the vertex is also a zero

b) The zero is at $(1,0)$

axis of symmetry: $x=himplies x=1$

Since $a=-5<0$, it opens downward.

c) The domain of quadratic functions are all set of real numbers.

domain: ${xinbold{R}}$

The range of quadratic function with vertex $(h,k)$ that opens downward is $f(x)leq k$

range: ${ f(x)inbold{R};|;f(x)leq 0}$

a) $f(x)=-5(x-1)^2$ , vertex$(1,0)$

b) zero at $x=1$, axis of symmetry $x=1$, opens down

c) domain = ${ xinbold{R}}$ ; range = ${ yinbold{R};|;yleq 0}$

d) see graph inside

a) $f(x)=-2x^2-8x+3$

This is in standard form $f(x)=ax^2+bx+c$

Since $a=-2<0$, it will open down so it has a maximum value.

The maximum value can be obtained from the vertex $(h,k)$

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

$k=f(h)$

In this case, $a=-2$ , $b=-8$ , $c=3$

$h=-dfrac{b}{2a}=-dfrac{-8}{2(-2)}=-2$

$k=f(-2)=-2(-2)^2-8(-2)+3=11$

The maximum value is $11$

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

This is in factored form $f(x)=a(x-r)(x-s)$

Since $a=3>0$, it will open up so it has a minimum.

The minimum value can be obtained from the vertex$(h,k)$

$h=dfrac{r+s}{2}$

$k=f(h)$

In this case, $r=1$, $s=-5$

$h=dfrac{1+(-5)}{2}=-2$

$k=f(-2)=3(-2-1)(-2+5)=-27$

The minimum value is $-27$

a) maximum value: use $h=-dfrac{b}{2a}$

b) minimum value: average the zeros

$bold{Concept:}$ Here are the forms of quadratic function and their readily available information.

vertex form: $y=a(x-h)^2+k$ $implies$vertex$(h,k)$
factored form: $y=a(x-r)(x-s)$ $implies$ zeros at $x=r$ and $x=s$

standard form: $y=ax^2+bx+c$ $implies$ $y$-intercept at $(0,c)$

$bold{Solution:}$

a) The vertex is readily available in vertex form.

b) The $y$-intercept is readily available in standard form.

c) The zeros are readily available in factored form

d) The axis of symmetry is $x=h$ which can be obtained from vertex form.

e) The domain is ${xinbold{R}}$ for quadratic functions so any form will do.

The range is readily available in vertex form.

if $a>0$ (open up) , range = ${ yinbold{R};|ygeq k}$

if $a<0$ (open down), range = ${ yinbold{R};|yleq k}$

a) vertex form ; vertex is readily available

b) standard form; $y$-intercept is visible

c) factored form; zeros are visible

d) vertex form; use $x$-coordinate of the vertex

e) vertex form ; use vertex and direction of opening

Let $x$ and $w$ be the length and width of the rectangle respectively.

The fencing corresponds to the perimeter of the rectangle which is

Perimeter = $2(x+w)$

Perimeter $=2400$

$2(x+w)=2400implies x+w=1200$

Express the width in terms of the length $x$

$w=1200-x$

Now, we want to maximize the area

Area = $xcdot w$
Area =$x(1200-x)$

This is quadratic equation in factored form. The value of $x$ that maximizes the area is at the vertex which is the average of two zeros $x=0$ and $x=1200$

$x=dfrac{0+1200}{2}=600$

Now, use this value of $x$ to find the area

$A=x(1200-x)$

$A=600(1200-600)$

$A=360,000$ m$^2$

Therefore, the maximum area is $360,000$ m$^2$

$360,000$ m$^2$

We can obtain the inverse of a function by swapping the positions of $x$ and $y$ and then solve for $y$

$y=2(x-1)^2-3$

Swap variables

$x=2(y-1)^2-3$

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

$(y-1)^2=dfrac{x+3}{2}$

$y-1=pmsqrt{dfrac{x+3}{2}}$

$y=1pmsqrt{dfrac{x+3}{2}}$

Therefore, the inverse of $f(x)$ is

$$
f^{-1}(x)=1pmsqrt{dfrac{x+3}{2}}
$$

$$
f^{-1}(x)=1pmsqrt{dfrac{x+3}{2}}
$$

Remember the following algebraic rules:

Distributive property:

$(a+b)(c+d)=a(c+d)+b(c+d)$

$a(c+d)=ac+ad$

Rule of radicals:

$sqrt{a}times sqrt{b}=sqrt{ab}$

$sqrt{a^2b}=asqrt{b}$

$$
ksqrt{a}+jsqrt{a}+msqrt{b}=(k+j)sqrt{a}+msqrt{b}
$$

a) $(2-sqrt{8})(3+sqrt{2})$

Apply distributive property

$=2(3+sqrt{2})-sqrt{8}(3+sqrt{2})$

$=6+2sqrt{2}-3sqrt{8}-sqrt{16}$

Factor the expression inside the radical such that one factor is a perfect square

$=6+2sqrt{2}-3sqrt{4times 2}-sqrt{4^2}$

Apply rule of radicals

$=6+2sqrt{2}-3cdot 2 sqrt{2}-4$

$=6+2sqrt{2}-6sqrt{2}-4$

$$
=2-4sqrt{2}
$$

b) $(3+sqrt{5})(5-sqrt{10})$

Apply distributive property

$=3(5-sqrt{10})+sqrt{5}(5-sqrt{10})$

Apply rule of radicals

$=15-3sqrt{10}+5sqrt{5}-sqrt{5times 10}$

$=15-3sqrt{10}+5sqrt{5}-sqrt{50}$

$=15-3sqrt{10}+5sqrt{5}-sqrt{25times 2}$

$=15-3sqrt{10}+5sqrt{5}-sqrt{5^2times 2}$

$=15-3sqrt{10}+5sqrt{5}-5sqrt{2}$

c) Radicals can only be combined if they have the same expression inside the radical sign. $sqrt{8}$ in part (a) can be simplified to $2sqrt{2}$ which can then be combined resulting in fewer terms. Whereas part (b) resulted in radicals that cannot be combined.

a) $2-4sqrt{2}$

b) $15-3sqrt{10}+5sqrt{5}-5sqrt{2}$

c) Radicals can only be combined if they have the same expression inside the radical sign. $sqrt{8}$ in part (a) can be simplified to $2sqrt{2}$ which can then be combined resulting in fewer terms, whereas part (b) resulted in radicals that cannot be combined

$bold{Concept:}$ A quadratic equation $f(x)=ax^2-bx+c$ has one root if

$$
b^2-4ac=0
$$

$bold{Solution:}$ In this case,

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

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

$16-4k^2=0$

$4k^2=16$

$k^2=4$

$k=pmsqrt{4}$

$k=pm 2$

Therefore, the value of $k$ that would lead to only one root is either
$k=2$ or $k=-2$

$k=-2$ or $2$

$bold{Concept:}$ The graph of a quadratic function $f(x)$ intersects with the line $g(x)$ if $f(x)-g(x)=0$ have at least one solution, that means $b^2-4acgeq 0$

In this case, $f(x)=2x^2-3x+2$ and $g(x)=6x-5$

$f(x)-g(x)=2x^2-3x+2-6x+5=0$

$2x^2-9x+7=0$

$b^2-4ac=(-9)^2-4(2)(7)=25>0$

Thus, it has two intersection points.

$(2x-7)(x-1)=0$

$x=dfrac{7}{2}$ or 1

To determine the intersection points, evaluate $g(x)$ or $f(x)$ using each value of $x$

$gleft(dfrac{7}{2}right)=6left(dfrac{7}{2}right)-5=16implies left(dfrac{7}{2},16right)$

$g(1)=6(1)-5=1implies (1,1)$

Therefore, $f(x)$ and $g(x)$ intersects at $left(dfrac{7}{2},16right)$ and $(1,1)$

Yes because $b^2-4acgeq 0$ for $f(x)-g(x)=0$.
Intersection points are $left(dfrac{7}{2},16right)$ and $(1,1)$

$bold{Concept:}$ To determine the equation of a parabola from a given graph, find the coordinate of the vertex $(h,k)$ then write the vertex form $y=a(x-h)^2+k$

Then use any point on the parabola to evaluate $a$

$bold{Solution:}$ From the graph, the vertex is $(4,3)$ and passes through $(3,2)$

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

Substitute the coordinates of $(3,2)$

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

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

Therefore, the equation of the parabola is

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

Now, we have to write in the standard form $y=ax^2+bx+c$

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

$y=-[x^2-8x+16]+3$

$y=-x^2+8x-16+3$

$$
y=-x^2+8x-13
$$

$$
y=-x^2+8x-13
$$