Answer:
the coordinates of the vertex are: (2.5, -24.5)
Step-by-step explanation:
Recall that when we have a quadratic in its standard form:
[tex]f(x)=ax^2+bx+c[/tex]
the position of the x-coordinate for the vertex can be obtained via:
[tex]x_{vertex}=\frac{-b}{2a}[/tex]
Then in order to find the vertex, first we write the expression in standard form:
[tex]f(x)=(x-6)(2x+2)\\f(x)=2x^2+2x-12x-12\\f(x)=2x^2-10x-12[/tex]
Now that we have the values for the parameters "[tex]a[/tex]" and "b" we find the x of the vertex:
[tex]x_{vertex}=\frac{-b}{2a}=\frac{10}{2*2}=\frac{5}{2}[/tex]
Now we use this x-value in the function to find the correspondent y-value of the vertex:
[tex]f(x)=2x^2-10x-12\\f(\frac{5}{2} )=2\,(\frac{5}{2} )^2-10(\frac{5}{2} )-12\\f(\frac{5}{2} )=-24.5[/tex]
Then, the coordinates of the vertex are: (2.5, -24.5)
