Answer:
SEE BELOW
Step-by-step explanation:
to understand this
you need to know about:
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:
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:
[tex]3 = a(1 {)}^{2} + 4[/tex]
simplify exponent:
[tex]3 = a + 4[/tex]
therefore
[tex]a = - 1[/tex]
our vertex form of quadratic equation is
let's simplify it to standard form
simplify square:
[tex]y = - ( {x}^{2} + 2x + 1) + 4[/tex]
simplify parentheses:
[tex]y = - {x}^{2} - 2x - 1 + 4[/tex]
simplify addition:
[tex]y = - {x}^{2} - 2x + 3[/tex]
therefore our answer is D)y=-x²-2x+3
the domain of the function
[tex]x\in \mathbb{R}[/tex]
and the range of the function is
[tex]y\leqslant 4[/tex]
zeroes of the function:
[tex] - {x}^{2} - 2x + 3 = 0[/tex]
[tex] \sf divide \: both \: sides \: by \: - 1[/tex]
[tex] {x}^{2} + 2x - 3 = 0[/tex]
[tex] \implies \: {x}^{2} + 3x - x + 3 = 0[/tex]
factor out x and -1 respectively:
[tex] \sf \implies \: x(x + 3) - 1(x + 3 )= 0[/tex]
group:
[tex] \implies \: (x - 1)(x + 3) = 0[/tex]
therefore
[tex] \begin{cases} x_{1} = 1 \\ x_{2} = - 3\end{cases}[/tex]
