Question 1028268: Write the standard form of the quadratic function that has the indicated vertex and whose graph passes through the given point. Use a graphing utility to verify your result.
Vertex:
(−6, −4)
Point:
(−4, 1)
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
The answer depends on what you mean by standard form. Some people call  standard form and ^2\ +\ k) vertex form. And some people refer to as standard form.
Either way there are two ways to go about this.
Since you are given the vertex and one other point, you can find a third point on the graph by considering symmetry. Since the point (-4,1) is two units to the right of the axis of symmetry, then the function value two units to the left of the axis must be the same, hence the point (-8,1) must also be on the graph.
Assuming as standard form, if (-4,1) is on the graph, then it must be true that:
Likewise, considering the other two points,
The solution to the 3X3 system of equations yields the standard form coefficients.
I'll leave it to you to verify that  ,  , and  .
On the other hand, if by standard form you mean what most folks call the vertex form, you can proceed as follows:
^2\ +\ k)
Substituting the vertex coordinates:
^2\ -\ 4)
Then, since you know that the point ) is on the graph,
^2\ -\ 4\ =\ 1)
And all you need to do is solve for  . Any doubt in your mind that the result will be ?
John
My calculator said it, I believe it, that settles it
|
|
|
