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Answer :

Semsee45go

Answer:

The vertex of the graph is:

  • the minimum point for a parabola that opens upwards
  • the maximum point for a parabola that opens downwards

As this parabola opens upwards, the vertex is the minimum point.

From inspection of the graph, vertex = (-3, -1)

Equation of the graph

From inspection we can see that the x-intercepts are when
x = -4 and x = -2

[tex]\sf x = -4 \implies x+4=0[/tex]

[tex]\sf x = -2 \implies x+2=0[/tex]

Therefore:

[tex]\sf y=(x+4)(x+2)[/tex]

Expand:

[tex]\sf \implies y=x^2+6x+8[/tex]

Nayefxgo

Answer:

[tex]B)f(x) = {x}^{2} + 6x + 8[/tex]

Step-by-step explanation:

Let [tex] f(x)=x^2[/tex] be the parent function. With transformation of function, firstly,we know that,

  1. We can move it up or down by adding a constant to the y-value

algebraically

  • g(x)=x²+C

Clearly, The parabola is moved down by 1 unit thus, C is -1. Therefore our function transforms to

  • f(x)=x²-1

secondly, we know that,

  • We can move it left or right by adding a constant to the x-value

algebraically,

  • g(x)=(x+C)²

in case,

  • C is positive, g(x) moves to the left and vise versa

Since the parabola is moved left by 3 unit, C is +3, and hence Our function eventually becomes

  • [tex]f(x) = (x + 3 {)}^{2} - 1[/tex]

simplifying it yields:

[tex]\implies \boxed{f(x) = {x}^{2} + 6x + 8}[/tex]

Hence,B is our required answer

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