[tex]f(x)=x^2-4x-12[/tex]
x-intercpets: value of x when f(x)=0. Use the quadratic formula:
[tex]\begin{gathered} x^2-4x-12=0 \\ \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \\ x=\frac{-(-4)\pm\sqrt[]{(-4)^2-4(1)(-12)}}{2(1)} \\ \\ x=\frac{4\pm\sqrt[]{64}}{2} \\ \\ x=\frac{4\pm8}{2} \\ \\ x_1=\frac{4+8}{2}=6 \\ \\ x_2=\frac{4-8}{2}=-2 \end{gathered}[/tex]X-intercept: (6,0) and (-2,0)
y-intercept: value of f(x) when x=0:
[tex]\begin{gathered} f(0)=0^2-4(0)-12 \\ f(0)=-12 \end{gathered}[/tex]Y-intercept: (0, -12)
Vertex: use the next foruma to find the x value in vertex:
[tex]\begin{gathered} x=-\frac{b}{2a} \\ \\ x=-\frac{(-4)}{2(1)} \\ \\ x=\frac{4}{2}=2 \end{gathered}[/tex]
Use that value of x to find the coordinate in y for the vertex:
[tex]\begin{gathered} f(2)=2^2-4(2)-12 \\ f(2)=4-8-12 \\ f(2)=-16 \end{gathered}[/tex]Vertex: (2, -16)
Axis of symmetry: value of x in the vertex.
Axis of simmetry: x=2
Maximim or minimum: if in the quadratic equation, the coefficient of the square x is possitive the parabola opens up and has a minimun value. If the coefficient of the square x is negative the parabola opens down and and has a maximum value.
Minimum
The minimum value is in the vertex (2 ,-16) is the value of the function f(x) in the vertex (the value of coordinate y):
Min value: -16Graph:
