Sagot :
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]
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,
- 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