How do you write a quadratic equation with x-intercepts: -3,2 ; point: (3,6)?

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How do you write a quadratic equation with x-intercepts: -3,2 ; point: (3,6)?

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Answer 1 · 2016-07-17T00:06:00

Use a couple of quadratic properties and algebra to find the equation is $y = x^{2} + x - 6$ $y=x^2+x-6$ .

Explanation:

If a quadratic equation has solutions $x = a$ $x=a$ and $x = b$ $x=b$ , then $x - a = 0$ $x-a=0$ and $x - b = 0$ $x-b=0$ . Furthermore, the quadratic can be written as $y = c (x - a) (x - b)$ $y=c(x-a)(x-b)$ , where $c$ $c$ is some constant. The reasoning is that if you set $y$ $y$ equal to $0$ $0$ , you get:
$c (x - a) (x - b) = 0$ $c(x-a)(x-b)=0$
Which is the same as:
$(x - a) (x - b) = 0$ $(x-a)(x-b)=0$
And so the solutions are $x = a$ $x=a$ and $x = b$ $x=b$ - which is exactly what we started with.

Alright, enough theory - let's get on with it! We are told that the $x$ $x$ -intercepts are $- 3$ $-3$ and $2$ $2$ , and since $x$ $x$ -intercepts are the same thing as zeros, $x = - 3$ $x=-3$ and $x = 2$ $x=2$ are solutions. Following the process from above, we can write the quadratic as:
$y = c (x + 3) (x - 2)$ $y=c(x+3)(x-2)$

To solve for $c$ $c$ , we use the other piece of info we were given: the point $(3, 6)$ $(3,6)$ :
$y = c (x + 3) (x - 2)$ $y=c(x+3)(x-2)$
$\to 6 = c (3 + 3) (3 - 2)$ $->6=c(3+3)(3-2)$
$\to 6 = c (6) (1)$ $->6=c(6)(1)$
$\to 6 = 6 c \to c = 1$ $->6=6c->c=1$

So the equation of the quadratic is:
$y = 1 (x + 3) (x - 2)$ $y=1(x+3)(x-2)$
$\to y = (x + 3) (x - 2) = x^{2} + 3 x - 2 x - 6 = x^{2} + x - 6$ $->y=(x+3)(x-2)=x^2+3x-2x-6=x^2+x-6$

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How do you write a quadratic equation with x-intercepts: -3,2 ; point: (3,6)?

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