Question

What would I set the scales to and where is the roots ,vertex and two other points

Answer 1

SOLUTION

Write out the equation

$y=x^2+4x-12$

The root of the equation is

$\begin{gathered} x^2+4x-12=0 \\ x^2+6x-2x-12=0 \\ x(x+6)-2(x+6)=0 \\ (x+6)(x-2)=0 \end{gathered}$

Equate the factors to zero

$\begin{gathered} x+6=0,x-2=0 \\ x=-6,x=2 \end{gathered}$

Hence, the root of the equation is

(-6,0) and (2,0)

The vertex of the equation will be obtained using

$\begin{gathered} u\sin g\text{ the genera form of a quadractic equation } \\ ax^2+bx+c=0,\text{ tye vertex is given by } \\ \: x_v=-\frac{b}{2a} \\ \text{where } \\ a=1,b=4,c=-12\text{ from the given equation } \end{gathered}$

Then

$\begin{gathered} x=-\frac{4}{2\times1}=-\frac{4}{2}=-2 \\ \text{Then} \\ y=x^2+4x-12,\text{ where x=-2} \\ y=^{}(-2)^2+4(-2)-12=4-8-12=-16 \\ \text{Vertex (-2,-16)} \end{gathered}$

Then

$\begin{gathered} \text{The other two point could be } \\ x=0,y=0^2+4(0)-12=0+0-12=-12\text{ } \\ (0,-12) \\ x=2,y=2^2+4(2)-12=4+8-12=0 \\ (2,0) \end{gathered}$

Plotting this point using the scale 2unit on both axes, the image of the graph will be