Question

Need help with question 8 Only Part B C and D

Answer 1

Step-by-step explanation:

Given:

`g(x)=3x^2+12x+9`

B.)

To find the vertex, we use the following formula:

`\begin{gathered} x=-(b)/(2a) \\ \text{From the Form:} \\ y=ax^2+bx+c \end{gathered}`

So based on the given equation, the values of a and b are:

a=3

b=12

We plug in what we know:

`\begin{gathered} x=-(b)/(2a) \\ =-(12)/(2(3)) \\ x=-2 \end{gathered}`

Next, we plug in x=-2 into g(x)=3x^2+12x+9:

`\begin{gathered} g(x)=3x^2+12x+9 \\ =3(-2)^2+12(-2)+9 \\ \text{Calculate} \\ g(x)=-3 \end{gathered}`

Therefore, the vertex is (-2,-3).

C.

Now, to find the axis of symmetry, we also use the formula x=-b/2a since it is the vertical line that goes through the vertex.

Therefore, the Axis of Symmetry for the given equation is x = -2.

D.

We let g(x)=0 to find the x-intercept:

`\begin{gathered} 3x^2+12x+9=0 \\ \text{Simplify} \\ =3(x^2+4x+3) \\ =3(x+1)(x+3) \end{gathered}`

Based on the factors, the values for x are:

x=-1

x=-3

Therefore, the x intercept points are:

(-1,0),(-3,0)

To get the y-intercept, we let x=0 and plug in into g(x)=3x^2+12x+9. So,

`\begin{gathered} g(x)=3x^2+12x+9 \\ g(0)=3(0)^2+12(0)+9 \\ g(0)=9 \end{gathered}`

Therefore, the y intercept point is (0,9).