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

`\displaystyle 2 {x}^(2) + 4x - 30 = 0`

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