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In this section, you will:
• Recognize characteristics of parabolas.
• Understand how the graph of a parabola is related to its quadratic function.
• Determine a quadratic function’s minimum or maximum value.
• Solve problems involving a quadratic function’s minimum or maximum value.

Curved antennas, such as the ones shown in [link] , are commonly used to focus microwaves and radio waves to transmit television and telephone signals, as well as satellite and spacecraft communication. The cross-section of the antenna is in the shape of a parabola, which can be described by a quadratic function.

In this section, we will investigate quadratic functions, which frequently model problems involving area and projectile motion. Working with quadratic functions can be less complex than working with higher degree functions, so they provide a good opportunity for a detailed study of function behavior.

## Recognizing characteristics of parabolas

The graph of a quadratic function is a U-shaped curve called a parabola . One important feature of the graph is that it has an extreme point, called the vertex    . If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value . In either case, the vertex is a turning point on the graph. The graph is also symmetric with a vertical line drawn through the vertex, called the axis of symmetry    . These features are illustrated in [link] .

The y -intercept is the point at which the parabola crosses the y -axis. The x -intercepts are the points at which the parabola crosses the x -axis. If they exist, the x -intercepts represent the zeros     , or roots    , of the quadratic function, the values of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ at which $\text{\hspace{0.17em}}y=0.$

## Identifying the characteristics of a parabola

Determine the vertex, axis of symmetry, zeros, and $\text{\hspace{0.17em}}y\text{-}$ intercept of the parabola shown in [link] .

The vertex is the turning point of the graph. We can see that the vertex is at $\text{\hspace{0.17em}}\left(3,1\right).\text{\hspace{0.17em}}$ Because this parabola opens upward, the axis of symmetry is the vertical line that intersects the parabola at the vertex. So the axis of symmetry is $\text{\hspace{0.17em}}x=3.\text{\hspace{0.17em}}$ This parabola does not cross the $\text{\hspace{0.17em}}x\text{-}$ axis, so it has no zeros. It crosses the $\text{\hspace{0.17em}}y\text{-}$ axis at $\text{\hspace{0.17em}}\left(0,7\right)\text{\hspace{0.17em}}$ so this is the y -intercept.

## Understanding how the graphs of parabolas are related to their quadratic functions

The general form of a quadratic function presents the function in the form

$f\left(x\right)=a{x}^{2}+bx+c$

where $\text{\hspace{0.17em}}a,b,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ are real numbers and $\text{\hspace{0.17em}}a\ne 0.\text{\hspace{0.17em}}$ If $\text{\hspace{0.17em}}a>0,\text{\hspace{0.17em}}$ the parabola opens upward. If $\text{\hspace{0.17em}}a<0,\text{\hspace{0.17em}}$ the parabola opens downward. We can use the general form of a parabola to find the equation for the axis of symmetry.

The axis of symmetry is defined by $\text{\hspace{0.17em}}x=-\frac{b}{2a}.\text{\hspace{0.17em}}$ If we use the quadratic formula, $\text{\hspace{0.17em}}x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a},\text{\hspace{0.17em}}$ to solve $\text{\hspace{0.17em}}a{x}^{2}+bx+c=0\text{\hspace{0.17em}}$ for the $\text{\hspace{0.17em}}x\text{-}$ intercepts, or zeros, we find the value of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ halfway between them is always $\text{\hspace{0.17em}}x=-\frac{b}{2a},\text{\hspace{0.17em}}$ the equation for the axis of symmetry.

[link] represents the graph of the quadratic function written in general form as $\text{\hspace{0.17em}}y={x}^{2}+4x+3.\text{\hspace{0.17em}}$ In this form, $\text{\hspace{0.17em}}a=1,b=4,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}c=3.\text{\hspace{0.17em}}$ Because $\text{\hspace{0.17em}}a>0,\text{\hspace{0.17em}}$ the parabola opens upward. The axis of symmetry is $\text{\hspace{0.17em}}x=-\frac{4}{2\left(1\right)}=-2.\text{\hspace{0.17em}}$ This also makes sense because we can see from the graph that the vertical line $\text{\hspace{0.17em}}x=-2\text{\hspace{0.17em}}$ divides the graph in half. The vertex always occurs along the axis of symmetry. For a parabola that opens upward, the vertex occurs at the lowest point on the graph, in this instance, $\text{\hspace{0.17em}}\left(-2,-1\right).\text{\hspace{0.17em}}$ The $\text{\hspace{0.17em}}x\text{-}$ intercepts, those points where the parabola crosses the $\text{\hspace{0.17em}}x\text{-}$ axis, occur at $\text{\hspace{0.17em}}\left(-3,0\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(-1,0\right).$

Cos45/sec30+cosec30=
Cos 45 = 1/ √ 2 sec 30 = 2/√3 cosec 30 = 2. =1/√2 / 2/√3+2 =1/√2/2+2√3/√3 =1/√2*√3/2+2√3 =√3/√2(2+2√3) =√3/2√2+2√6 --------- (1) =√3 (2√6-2√2)/((2√6)+2√2))(2√6-2√2) =2√3(√6-√2)/(2√6)²-(2√2)² =2√3(√6-√2)/24-8 =2√3(√6-√2)/16 =√18-√16/8 =3√2-√6/8 ----------(2)
exercise 1.2 solution b....isnt it lacking
I dnt get dis work well
what is one-to-one function
what is the procedure in solving quadratic equetion at least 6?
Almighty formula or by factorization...or by graphical analysis
Damian
I need to learn this trigonometry from A level.. can anyone help here?
yes am hia
Miiro
tanh2x =2tanhx/1+tanh^2x
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)=cotb ... pls some one should help me with this..thanks in anticipation
f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)= cotb
favour
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
i am in
Cliff
hii
Amit
how are you
Dorbor
well
Biswajit
can u tell me concepts
Gaurav
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
helo
Akash
hlo
Akash
Hello
Hudheifa
which of these functions is not uniformly continuous on 0,1