a quadratic has zeros of -2 and 12. it has a stretch factor of -2. find the eqation in factored form and in standard form

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The quadratic has zeros of -2 and 12. So (x+2) and (x-12) are the factors of the equation.

(x+2) (x-12)=0

Again it has a stretch factor (the leading coefficient 'a') of -2. Hence the equation in factored form will be

a(x+2) (x-12)=0

`rArr` -2(x+2) (x-12)=0

`rArr` (-2x-4)(x-12)=0

`rArr` -2x^2+24x-4x+48=0

`rArr` -2x^2+20x+48=0

`rArr` 2x^2-20x-48=0

**This is the required quadratic equation.**

**Sources:**

A quadratic in factored form is in the form y=a(x-p)(x-q); p and q are the zeros of the function (roots, solutions, etc...) and a is the vertical stretch factor. (If a<0 then the graph is reflected over the horizontal line.)

So given the zeros -2 and 12 and a stretch factor of -2:

`y=-2(x+2)(x-12)`

** Note the sign change; it is (x-p), so if p=-2 then x-(-2)=x+2**

Standard form will be `y=ax^2+bx+c` ; we can expand the product.

`y=-2[x^2-10x-24]`

`y=-2x^2+20x+48`

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The factored form is y=-2(x+2)(x-12)

The standard form is `y=-2x^2+20x+48`

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