Question

Axis of sym: x =

Vertex: (
,
)

Do not do the domain

Range: Pick one: A. y≥4 B. y<4 C. y≤4 D. y≤−2 E. y<−2

The Range is

Zeros:
and


Equation:

Answer 1

Answer:

SEE BELOW

Step-by-step explanation:

to understand this

you need to know about:

• quadratic function
• PEMDAS

let's solve:

vertex:(h,k)

therefore

vertex:(-1,4)

axis of symmetry:x=h

therefore

axis of symmetry:x=-1

• to find the quadratic equation we need to figure out the vertex form of quadratic equation and then simply it to standard form i.e ax²+bx+c=0

vertex form of quadratic equation:

• y=a(x-h)²+k

therefore

• y=a(x-(-1))²+4
• y=a(x+1)²+4

it's to notice that we don't know what a is

therefore we have to figure it out

the graph crosses y-asix at (0,3) coordinates

so,

3=a(0+1)²+4

simplify parentheses:

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

simplify exponent:

$3 = a + 4$

therefore

$a = - 1$

our vertex form of quadratic equation is

• y=-(x+1)²+4

let's simplify it to standard form

simplify square:

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

simplify parentheses:

$y = - {x}^{2} - 2x - 1 + 4$

simplify addition:

$y = - {x}^{2} - 2x + 3$

therefore our answer is D)y=-x²-2x+3

the domain of the function

$x\in \mathbb{R}$

and the range of the function is

$y\leqslant 4$

zeroes of the function:

$- {x}^{2} - 2x + 3 = 0$

$\sf divide \: both \: sides \: by \: - 1$

${x}^{2} + 2x - 3 = 0$

$\implies \: {x}^{2} + 3x - x + 3 = 0$

factor out x and -1 respectively:

$\sf \implies \: x(x + 3) - 1(x + 3 )= 0$

group:

$\implies \: (x - 1)(x + 3) = 0$

therefore

$\begin{cases} x_{1} = 1 \\ x_{2} = - 3\end{cases}$