Question

I need help with number 5 the one in black

Answer 1

Given: The function below

`h(x)=-3x^2`

To Determine: The Vertex and the axis of symmetry

Solution

The vertex is given by the formula for a parabola as

`Vertex:(-(b)/(2a),f(-(b)/(2a)))`

From function given, let us determine the value of a, b, and c

`\begin{gathered} General\text{ equation is} \\ f(x)=ax^2+bx+c \\ Compare\text{ with the given equation} \\ h(x)=-3x^2 \\ a=-3,b=0,c=0 \end{gathered}`

Therefore

`\begin{gathered} Vertex=((-0)/(2*-3),h((-0)/(2*-3)) \\ Vertex=((0)/(6),h((0)/(6)) \\ Vertex=(0,h(0)) \end{gathered}`
`\begin{gathered} If,h(x)=-3x^2 \\ h(0)=-3(0)^2=-3(0)=0 \end{gathered}`

Hence, the vertex is (0, 0)

(b) The axis of symmetry is calculated using

`x=-(b)/(2a)`

Substitute the for a and b

`\begin{gathered} a=-3,b=0 \\ x=-(0)/(2*-3) \\ x=(0)/(6) \\ x=0 \end{gathered}`

Hence, the axis of symmetry is x = 0

The maximum point of the graph is 0. This is shown in the graph of the function below