Question

how to write the equation with (-2,0) and (4,0) as the x intercepts and passes through the point (1,-27)

Answer 1

Given:

Two intercept points be (-2,0) and (4,0) and passing through the point (1,-27).

Let the equation of parabola be

$y=ax^2+bx+c$

Substituting the points we get,

$\begin{gathered} 4a-2b+c=0\ldots\text{ (1)} \\ 16a+4b+c=0\ldots\text{ (2)} \\ a+b+c=-27\ldots\text{ (3)} \end{gathered}$

Solving (1) and (2) we get,

$\begin{gathered} 12a+6b=0 \\ 2a+b=0\ldots\text{ (4)} \end{gathered}$

Solving (2) and (3) we get,

$\begin{gathered} 15a+3b=27 \\ 5a+b=9\ldots\text{ (5)} \end{gathered}$

Solving (4) and (5)

$\begin{gathered} 3a=9 \\ a=3 \end{gathered}$

Substituting a=3 in (4)

$\begin{gathered} 2(3)+b=0 \\ b=-6 \end{gathered}$

Substitute a=3 & b=-6 in (1)

$\begin{gathered} 3-6+c=-27 \\ -3+c=-27 \\ c=-24 \end{gathered}$

Equation of parabola is

$y=3x^2-6x-24$