Question

All you need is in the photo and it's homework please don't do step by step

Answer 1

`f(x)=x^2-4x-12`

x-intercpets: value of x when f(x)=0. Use the quadratic formula:

`\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=(4\pm8)/(2) \\ \\ x_1=(4+8)/(2)=6 \\ \\ x_2=(4-8)/(2)=-2 \end{gathered}`X-intercept: (6,0) and (-2,0)

y-intercept: value of f(x) when x=0:

`\begin{gathered} f(0)=0^2-4(0)-12 \\ f(0)=-12 \end{gathered}`Y-intercept: (0, -12)

Vertex: use the next foruma to find the x value in vertex:

`\begin{gathered} x=-(b)/(2a) \\ \\ x=-((-4))/(2(1)) \\ \\ x=(4)/(2)=2 \end{gathered}`

Use that value of x to find the coordinate in y for the vertex:

`\begin{gathered} f(2)=2^2-4(2)-12 \\ f(2)=4-8-12 \\ f(2)=-16 \end{gathered}`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: