Question

Find a formula for a call dragon function with 0 at X =3 and x=-5

Answer 1

quadratic function

zeros: x = -5 and 3

y-intercept at (0,12)

Since, the factors are:

(x - (-5) )

(x - 3)

We write the function as it follows

`f(x)=a(x+5)(x-3)`

we can find a from a given point, for example (0,12)

`\begin{gathered} f(0)=a\cdot(0+5)(0-3)=12 \\ a\cdot5\cdot(-3)=12 \\ a=-0.8 \end{gathered}`

therefore, the function is

`f(x)=-0.8(x+5)(x-3)`

we can rewrite the expression as it follows:

`\begin{gathered} -0.8\mleft(x+5\mright)\mleft(x-3\mright) \\ =-0.8\mleft(x^2+5x-3x-15\mright) \\ =-0.8\mleft(x^2+2x-15\mright) \\ =-0.8x^2-0.8\cdot\: 2x-0.8\mleft(-15\mright) \end{gathered}`

simplify

`=-0.8x^2-1.6x+12`