A quadratic equation is of the form below:
[tex]ax^2+bx+c=y[/tex]
The vertex of the quadratic equation is the value of y where the curve cut the y-axis.
To find the vertex, we would first first find the x-coordinate of the equation using
[tex]x=-\frac{b}{2a}[/tex]
To find the vertex, we would substitute for x in the equation to get y.
Given the quadratic function below
[tex]f(x)=-3x^2-4x+7[/tex]
It can be observed that
[tex]a=-3;b=-4,c=7[/tex]
The x-coordinate of the vertex is as shown below:
[tex]\begin{gathered} x=-\frac{b}{2a} \\ x=\frac{--4}{2(-3)} \\ x=\frac{4}{-6} \\ x=-\frac{2}{3} \end{gathered}[/tex]
The vertex would be
[tex]\begin{gathered} f(-\frac{2}{3})=-3(-\frac{2}{3})^2-4(-\frac{2}{3})+7 \\ f(-\frac{2}{3})=-3(\frac{4}{9})+\frac{8}{3}+7 \\ f(-\frac{2}{3})=-\frac{4}{3}+\frac{8}{3}+7 \\ f(-\frac{2}{3})=\frac{-4+8}{3}+7 \\ f(-\frac{2}{3})=\frac{4}{3}+7 \\ f(-\frac{2}{3})=1\frac{1}{3}+7=8\frac{1}{3}=\frac{25}{3} \end{gathered}[/tex]
Hence, the vertex of the quadratic function is (-2/3,25/3)
It should be noted that when a is positive, the quadratic graph opens upward and the vertex is minimum but when a is negative, the quadratic curve opens downward, and the vertex would be maximum
The graph of the quafratic function given is shown below
It can be observed that the vertex is maximum
