Given:
The quadratic equation is
[tex]y=3x^2-18x+15[/tex]
To find:
The x-coordinate and y-coordinate of the vertex.
Solution:
If a quadratic function is defined by [tex]f(x)=ax^2+bx+c[/tex], then the vertex is defined as:
[tex]Vertex=\left(\dfrac{-b}{2a},f(\dfrac{-b}{2a})\right)[/tex]
If [tex]a<0[/tex], then the function has a maximum at the vertex and ff [tex]a>0[/tex], then the function has a minimum at the vertex.
We have,
[tex]y=3x^2-18x+15[/tex]
Here, [tex]a=3,b=-18,c=15[/tex].
Since [tex]a=3>0[/tex], therefore the function has a minimum at the vertex. So, fill min in first blank.
Now,
[tex]\dfrac{-b}{2a}=\dfrac{-(-18)}{2(3)}[/tex]
[tex]\dfrac{-b}{2a}=\dfrac{18}{6}[/tex]
[tex]\dfrac{-b}{2a}=3[/tex]
Putting x=3 in the given function, we get
[tex]y=3(3)^2-18(3)+15[/tex]
[tex]y=27-54+15[/tex]
[tex]y=-12[/tex]
Therefore, the vertex is at point (3,-12).
