Question

Determine an equation for the pictured graph. Write your answer in factored form and assume the leading coefficient is either 1 or -1, you should be able to determine which is the case by looking at the graph.

Answer 1

Ok so here we need to find a suitable equation for this graph. Since we are being asked for a factored form the equation is a polynomial one:

`y=a\cdot(x-r_1)^A\cdot(x-r_2)^B\cdot(x-r_3)^C\ldots`

Where the r terms represent the roots of the equation which are the points where there are x-intercepts. So first let's identify these points. The graph has three x-intercepts: x=-2, x=0 and x=1 so we have:

`\begin{gathered} y=a\cdot(x-(-2))^A\cdot(x-0_{})^B\cdot(x-1_{})^C \\ y=a\cdot(x+2)^A\cdot x^B\cdot(x-1_{})^C \end{gathered}`

To find the exponents A, B and C we need to look at the sign of the graph near each x-intercept. If in a given x-intercept the graph changes its sign then the exponent is an odd number. If it doesn't then its an even number.

In the case of x-intercept at x = -2 you can see that the graph changes from negative values to positive values so the exponent A has to be an odd number, let's choose 1. On x=0 the graph sign changes again, it's positive at the left and negative at the right so B is also an odd number and we'll choose number 1 again. Finally on x=1 the graph is negative on both sides of the intercept so C is an even number, let's say 2. All in all we have A=B=1 and C=2. Then:

`\begin{gathered} y=a\cdot(x+2)^1\cdot x^1\cdot(x-1_{})^2 \\ y=a\cdot(x+2)^{}\cdot x^{}\cdot(x-1_{})^2 \end{gathered}`

So we just need to find a. You can notice that for very large values of x (for x>1) the value of the function must be negative. So let's take a random value ok x that is bigger than 1 let's say 10:

`\begin{gathered} y=a\cdot(10+2)^{}\cdot10^{}\cdot(10-1_{})^2 \\ y=a9720 \end{gathered}`

So we have y=9720*a and according to what I stated before this value must be negative. 9720 is a positive number so "a" has to be negative and since the two possible options for a are 1 or -1 then we have that a=-1. Then:

`y=(-1)\cdot(x+2)^{}\cdot x^{}\cdot(x-1_{})^2`