Question

A quadratic relation has zeros at 3 and –5. The parabola goes through the point (2, –14). What is the equation of the parabola in standard form?

Answer 1

Answer:

$\displaystyle 2 {x}^{2} + 4x - 30 = 0$

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

$\displaystyle \: y = a(x - x_{1})(x - x_{2})$

where a is the leading coefficient and x1 and x2 are the roots

substitute:

$\displaystyle y = a(x - (3))(x - ( - 5))$

simplify:

$\displaystyle y = a(x - 3)(x + 5)$

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:

$\displaystyle a(2 - 3)(2 + 5) = - 14$

simplify parentheses:

$\displaystyle a( - 1)(7) = - 14$

simplify multiplication:

$\displaystyle - 7a = - 14$

divide both sides by -7:

$\displaystyle a = 2$

altogether substitute:

$\displaystyle y = 2(x - 3)(x + 5)$

since it want the equation y should be

$\displaystyle 2(x - 3)(x + 5) = 0$

recall quadratic equation standard form:

$\displaystyle {ax}^{2} + bx + c = 0$

so simplify parentheses:

$\displaystyle 2( {x}^{2} + 2x - 15 ) = 0$

distribute:

$\displaystyle 2 {x}^{2} + 4x - 30 = 0$

hence,

the equation of the parabola in standard form is 2x²+4x-30=0