The line of symmetry of a horizontal parabola is a horizontal line, the equation of which takes the form $\displaystyle y = b$ for some $\displaystyle b$. The line of symmetry passes through the vertex, which here is $\displaystyle (8,4)$, so the equation must be $\displaystyle y = 4$.

The $\displaystyle y$-coordinate of the $\displaystyle y$-intercept of the graph of a function of the form $\displaystyle f(x)= Ax^{2}+Bx+C$ - a quadratic function - is the point $\displaystyle (0, f(0))$. Since $\displaystyle f(x)= A \cdot 0^{2}+B \cdot 0 +C = C$, the $\displaystyle y$-intercept is at the point $\displaystyle (0,C)$.

Because of this, the graph of $\displaystyle f(x) = 2x^{2}+ 4x+7$ has its $\displaystyle y$-intercept at $\displaystyle (0,7)$. Among the other choices, only $\displaystyle g(x) = 4x^{2}+ 8x+7$ has a graph with its $\displaystyle y$-intercept also at $\displaystyle (0,7)$.

The correct answer for the vertex is found by first finding the x of the vertex: 

$\displaystyle x=\frac{-b}{2a}\ where\ a=-1\ and\ b=2$

Plug in a and b to get: 

$\displaystyle x=\frac{-2}{2(-1)}=\boldsymbol{\mathit{}1}$

To find the y value of the vertex, plug in what was found for x above in the original f(x). 

$\displaystyle f(x)=-x^{2}+2x-8\ plug\ in\ x=1$

$\displaystyle f(1)=-(1)^{2}+2(1)-8=\boldsymbol{\mathit{}-7}$

a common mistake here is the order of operations, at the beginning the 1 is squared before it is multipled by the negative out front. 

$\displaystyle The\ vertex\ is\ \boldsymbol{\mathbf{\mathit{}}(1,-7)}$

Now we must consider if the vertex is a MAX or a MIN

Since the a value is negative, this means the parabola will open down, which means the vertex is the highest point on the graph. 

$\displaystyle The\ vertex\ is\ the\ \boldsymbol{MAX}$

$\displaystyle The\ -3\ at\ the\ beginning\ of\ the\ function\ determines\ if\ the\ graph\ opens\ up\ or\ down.$

$\displaystyle The\ parabola\ will\ open\ up\ if\ this\ value\ is\ positive.$

$\displaystyle The\ parabola\ will\ open\ down\ if\ this\ value\ is\ negative.$

$\displaystyle This\ parabola\ opens\ down\ which\ means\ the\ y-value\ of\ the\ vertex\ is\ the\ highest\ point.$

$\displaystyle This\ quadratic\ is\ in\ vertex\ form: f(x)=a(x-h)^{2}+k$

$\displaystyle The\ k\ value\ is\ the\ y-value\ of\ the\ vertex\ which\ is\ 7.$

$\displaystyle Given\ that\ the\ parabola\ opens\ down\ and\ the\ y-value\ of\ the\ vertex\ is\ 7,$

$\displaystyle the\ RANGE\ must\ be\ \mathbf{y\leq 7}.$

$\displaystyle The\ range\ for\ a\ quadratic\ is\ determined\ from\ the\ y-value\ of\ the\ vertex.$

$\displaystyle This\ equation\ is\ written\ in\ vertex\ form: y=a(x-h)^{2}+k$

$\displaystyle The\ k -value\ is\ the\ y-coordinate\ for\ the\ vertex.$

$\displaystyle For\ this\ function\ y=5(x+3)^{2}-4,\ the\ k\ value\ is -4.$

$\displaystyle We\ can\ also\ see\ from\ the\ a-value,\ which\ is\ positive\ 5,\ that\ this\ parabola\ will\ open\ up\ on\ the\ graph.$

$\displaystyle \\Both\ of\ these\ together\ give\ us\ the\ range\ which\ is\ y\geq-4.\ It\ is \geq\ because\ the\ parabola\ opens\ up.$

$\displaystyle To\ find\ the\ y-intercept,\ you\ must\ plug\ in\ x=0\ to\ the\ original\ equation.$

$\displaystyle y=5(0+3)^{2}-4$

 $\displaystyle Pay\ attention\ to\ order\ of\ operations\ by\ squaring\ the\ (-3)\ before\ multiplying\ by\ 5.$

$\displaystyle After\ simplifying\ we\ get\ y=41\ which\ is\ the\ y-intercept.$

When you look at the graph, you will see the x-intercepts are

$\displaystyle (4,0)(-3,0)$ 

and the y-intercept is 

$\displaystyle (0,-12)$.

These numbers are the solutions to the equation.

You can work backwards and see what the actual equation will come out as,

$\displaystyle y=(x-4)(x+3)$.

This would distribute to 

$\displaystyle y=x^2-4x+3x-12$ 

and then simplify to 

$\displaystyle y=x^2-x-12$.

This also would show a y-intercept of $\displaystyle (0,-12)$.

$\displaystyle Use \frac{-b}{2a}\ to\ calculate\ the\ x-value\ of\ the\ vertex.$

$\displaystyle \small Given\ the\ function\ f(x)=4x^{2}-16x+11\ is\ in\ standard\ form\ y=ax^{2}+bx+c$

$\displaystyle \small We\ can\ see\ that\ a=4, b=-16\ and\ c=11$

$\displaystyle \small Substitute\ these\ values\ in\ for\ x=\frac{-b}{2a}\ to\ get\ x=\frac{16}{8}=2$

$\displaystyle \small Now\ substitute\ x=2\ into\ the\ function\ f(x)\ to\ get:$

$\displaystyle \small f(2)=4(2)^{2}-16(2)+11=-5$

$\displaystyle \small The\ coordinates\ of\ the\ vertex\ are\ (2,-5)$

When finding the domain and range of a quadratic function, we must first find the vertex. 

$\displaystyle To\ find\ the\ vertex\ of\ a\ quadratic\ we\ must\ first\ find\ the\ x-coordinate\ using\ the\ formula:$

$\displaystyle x=\frac{-b}{2a}$

$\displaystyle We\ are\ given\ f(x)=-3x^2$

$\displaystyle This\ gives\ us\ an\ a-value\ of\ -3\ while\ b\ and\ c\ both\ equal\ 0.$

$\displaystyle Plugging\ these\ values\ in\ give\ us\ an\ x-coordinate\ and\ y-coordinate\ of\ 0.$

$\displaystyle The\ domain\ of\ all\ quadratics\ is\ ALL\ REAL\ NUMBERS\ (\mathbb{R})$

$\displaystyle With\ a\ negative\ a-value\ we\ know\ this\ quadratic\ opens\ down:$

$\displaystyle so\ the\ range\ is\ y\leq0$

The $\displaystyle x$-intercept(s) of the graph of $\displaystyle f(x)$ are the point(s) at which it intersects the $\displaystyle x$-axis. The $\displaystyle y$-coordinate of each is 0; their $\displaystyle x$-coordinate(s) are those value(s) of $\displaystyle x$ for which $\displaystyle f(x) = 0$, so set up, and solve for $\displaystyle x$, the equation:

$\displaystyle f(x)= 0$

$\displaystyle x^{2}-3x-10 = 0$

To solve this quadratic equation, first, factor the quadratic trinomial as 

$\displaystyle (x+a)(x+b)$

by finding two integers with sum $\displaystyle -3$ and product $\displaystyle -10$. From trial and error, we find the integers $\displaystyle -5$ and 2, so the equation can be rewritten as 

$\displaystyle (x-5)(x+2) = 0$

By the Zero Factor Principle, one of these two binomials is equal to 0, so either

$\displaystyle x-5 = 0$,

in which case 

$\displaystyle x = 5$,

or 

$\displaystyle x+2 = 0$,

in which case

$\displaystyle x= -2$.

The graph has two $\displaystyle x$-intercepts at the points $\displaystyle (-2, 0)$ and $\displaystyle (5, 0)$.

$\displaystyle f(x)$ is a polynomial function, and as such has the set of all real numbers as its domain.