I need help with this quadratic function… I thought I knew the answer, but obviously I don’t

Let us start with the following quadratic function:
[tex]f(x)=x^2-x-12[/tex]the X-intercepts are the collection of values to X which makes f(x) = 0, and it can be calculated by the Bhaskara formula:
[tex]x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]where the values a, b, and c are given by:
[tex]f(x)=ax^2+bx+c[/tex]Substituting the values from the proposed equation, we have:
[tex]\begin{gathered} x_{1,2}=\frac{1\pm\sqrt{1^2-4*1*(-12)}}{2*1} \\ x_{1,2}=\frac{1\pm\sqrt{1+48}}{2}=\frac{1\pm\sqrt{49}}{2} \\ x_{1,2}=\frac{1\pm7}{2} \\ \\ x_1=\frac{1+7}{2}=\frac{8}{2}=4 \\ x_2=\frac{1-7}{2}=-\frac{6}{2}=-3 \end{gathered}[/tex]From the above-developed solution, we are able to conclude that the solution for the first box is:
Now, the y-intercept, is just the value of y when x = 0, which can be calculated as follows:
[tex]\begin{gathered} f(0)=0^2-0-12=-12 \\ f(0)=-12 \end{gathered}[/tex]From this, we are able to conclude that the solution for the second box is:
Now, the vertex is the value of minimum, or maximum, in the quadratic equation, and use to be calculated as follows:
[tex]\begin{gathered} Vertex \\ x=-\frac{b}{2a} \\ y=\frac{4ac-b^2}{2a} \end{gathered}[/tex]substituting the values, we have:
[tex]\begin{gathered} x=-\frac{-1}{2*1}=\frac{1}{2} \\ y=\frac{4*1*(-12)-(-1)^2}{4*1}=\frac{-48-1}{4}=\frac{-49}{4} \end{gathered}[/tex]which means that the solution for the thirst box is:
Now, the line of symmetry equation of a quadratic function is a vertical line that passes through the vertex, which was calculated to be in the point: (1/2, -49,4).
Because this is a vertical line, it is represented as follows:
[tex]x=\frac{1}{2}[/tex]