Answer:
[tex] \displaystyle 2 {x}^{2} + 4x - 30 = 0[/tex]
Step-by-step explanation:
we are given the zeros and a point where it goes through of a quadratic equation
remember that when the roots are given then the function should be
[tex] \displaystyle \: y = a(x - x_{1})(x - x_{2}) [/tex]
where a is the leading coefficient and x1 and x2 are the roots
substitute:
[tex] \displaystyle y = a(x - (3))(x - ( - 5)) [/tex]
simplify:
[tex] \displaystyle y = a(x - 3)(x + 5) [/tex]
now the given point tells us that when x is 2 y is -14 therefore by using the point we can figure out a
substitute:
[tex] \displaystyle a(2 - 3)(2 + 5) = - 14[/tex]
simplify parentheses:
[tex] \displaystyle a( - 1)(7) = - 14[/tex]
simplify multiplication:
[tex] \displaystyle - 7a = - 14[/tex]
divide both sides by -7:
[tex] \displaystyle a = 2[/tex]
altogether substitute:
[tex] \displaystyle y = 2(x - 3)(x + 5) [/tex]
since it want the equation y should be
[tex] \displaystyle 2(x - 3)(x + 5) = 0[/tex]
recall quadratic equation standard form:
[tex] \displaystyle {ax}^{2} + bx + c = 0[/tex]
so simplify parentheses:
[tex] \displaystyle 2( {x}^{2} + 2x - 15 ) = 0[/tex]
distribute:
[tex] \displaystyle 2 {x}^{2} + 4x - 30 = 0[/tex]
hence,
the equation of the parabola in standard form is 2x²+4x-30=0
