SOLUTION: What is the solution for the quadratic equation depicted in the graph? I linked the image. https://i.imgur.com/QgAV6kb.gif I just can't get these through my head, no matter w

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: What is the solution for the quadratic equation depicted in the graph? I linked the image. https://i.imgur.com/QgAV6kb.gif I just can't get these through my head, no matter w      Log On


   

Question 1164935: What is the solution for the quadratic equation depicted in the graph? I linked the image.
https://i.imgur.com/QgAV6kb.gif
I just can't get these through my head, no matter what I do!



Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


In factored form, a quadratic equation is of the form

y+=+a%28x-p%29%28x-q%29

If x is equal to either p or q, then the value of y is zero, because one of the factors (x-p) or (x-q) is zero.

The values p and q -- that make the y value zero -- are called the roots of the equation.

In a graph, that means the graph crosses the x-axis at x=p and at x=q.

In this example, the graph crosses the x-axis at x=-3 and at x=-4. So the quadratic equation is of the form

y+=+a%28x-%28-3%29%29%28x-%28-4%29%29

or

y+=+a%28x%2B3%29%28x%2B4%29

What is left to do is to determine the value of the coefficient a.

To do that, use any point on the graph other than an x-intercept.

In this example, it looks as if y=2 when x=-2. So plug those coordinates into the formula and solve for a.

2+=+a%28%28-2%29%2B3%29%28%28-2%29%2B4%29
2+=+a%281%29%282%29
a+=+1

So the equation for the graph is

y+=+1%28x%2B3%29%28x%2B4%29

or just

y+=+%28x%2B3%29%28x%2B4%29