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

$Vertex:(-\frac{b}{2a},f(-\frac{b}{2a}))$

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=(\frac{-0}{2\times-3},h(\frac{-0}{2\times-3}) \\ Vertex=(\frac{0}{6},h(\frac{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

$x=-\frac{b}{2a}$

Substitute the for a and b

$\begin{gathered} a=-3,b=0 \\ x=-\frac{0}{2\times-3} \\ x=\frac{0}{6} \\ x=0 \end{gathered}$

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